^iffrrf^ffl-irr 


IN  MEMORIAM 
FLORIAN  CAJORl 


7rfk    .  IV 


This  book  is  intended  as  a  sequel  to  the  "  First  lessons 
in  Geometry,"  and,  therefore,  presupposes  some  acquaint- 
ance with  that  little  treatise.  I  think  it  better,  however, 
that  some  interval  should  elapse  between  the  study  of 
that  book  and  of  this,  —  during  which  time  the  child  may 
be  occupied  in  the  study  of  Arithmetic. 

Geometrical  facts  and  conceptions  are  easier  to  a  child 
than  those  of  Arithmetic,  but  arithmetical  reasoning  is 
easier  than  geometrical.  The  true  scientific  order  in  a 
mathematical  education  would  therefore  be,  to  begin  with 
the  facts  of  Geometry,  then  take  both  the  facts  and  rea- 
soning of  Arithmetic,  and  afterwards  return  to  Geometry, 
not  to  its  facts  only,  but  to  its  proofs. 

The  object  of  "First  Lessons  in  Geometry"  is  to  develop 
the  child's  powers  of  imagination  ;  the  object  of  this  book 
is  to  develop  his  powers  of  reasoning.  That  book  I  con- 
sider adapted  to  children  from  six  to  twelve  years  of  age, 
this  to  children  from  thirteen  to  eighteen  years  old. 

(3) 


918241 


CONTENTS. 

PART    I. 
CHAPTER   I. 

PAGE 

Prelbiinary, 11 

1,  Geometry  the  science  of  form.  2,  Imagination.  3,  Reasoning. 
4|  Application, 

CHAPTER   II. 
Definitions, 12 

6,  Geometry.  6,  Point.  7,  Line.  8,  Surface.  9,  Solid.  10,  Straight  line. 
11,  Curve.  12,  Plane.  1.3-14,  Angle.  15,  Parallels.  16,  Triangle.  17, 
Right  triangle.    18,  Parallelogram.    19,  Redtangle.    20,  Square. 

CHAPTER   III. 
Reasoning,    • 15 

21,  Experiment.  22,  Doubt.  23,  Error.  24,  No  error  small.  25,  To 
avoid  error.  20-30,  Sum  of  angles  in  a  triangle.  31,  Generality.  32, 
Analysis.  33,  Synthesis.  34,  Vertical  angles.  .35,  Alternate  internal 
angles.  36,  Proof  of  Art.  26.  37,  Experimental  proof.  38,  Analysis. 
39,  Synthesis.    40,  Reasoning. 

CHAPTER   IV. 
Analysis  and  Synthesis, 19 

41,  Synthesis.  42-44,  Must  follow  analysis.  45-47,  Reasoning  is  in- 
sight. 48-49,  Assumptions  to  be  avoided.  50,  Axioms.  61,  Inevitable 
inferences.    62,  Illustrated  by  a  string.    63,  Aristotle's  dictum. 

1*  (5) 


6  CONTENTS. 

CHAPTER  V. 
Variety  of  Paths,     . 24 

54-5C,  Various  proofs  of  same  truth.  67,  58,  Two  proofs  of  Art.  20.  59, 
60,  Algebraic  proof  of  Art.  20.  01,  02,  Kinematic  proofs  of  Art.  20.  63, 
All  proofs  useful.  —  Examples.  I.  External  internal  angles  equal. 
II.  If  angles  equal,  tlie  linos  parallel.  III.,  IV.,  V.,  Idem.  VI.  Paral- 
lels do  not  meet.    VII.  Two  perpendiculars  from  one  point  impossible. 

CHAPTER   VI. 
The  Pythagorean  Proposition, 28 

64-75,  Analysis  of  Pythagorean  proposition.  70,  Ratio.  77,  Its  nota- 
tion. 78,  Ratio  unchanged  by  multiplying  both  terms.  79,  Proportion. 
80,  Extremes  and  means.  81,  Product  of  means  equals  that  of  ex- 
tremes. 82,  Mean  proportional.  83,  Equals  square  root  of  product  of 
extremes.  84,  85,  Units.  80,  Area.  87,  Straight  line  makes  same  angle 
witli  parallels.  88,  89,  If  same  angles,  the  lines  are  parallel.  90,  Coinci- 
dence by  Buperimposition. 

CHAPTER   VII. 
The  Pythagorean  Proposition,  (continued),    .    .    33 

91,  One  side  and  adjacent  angles  given,  gives  the  triangle.  92, 
Parallelogram  has  its  opposite  sides  equal.  93,  Rotation  at  right 
angles  to  perpendicular  on  axis.  94,  95,  The  three  sides  given 
gives  the  triangle.  90,  Quadrangle  with  equal  opposite  sides  is  a 
parrallelogram.  97,  Area  of  rectangle.  98,  Infinitesimals  neglected. 
99, 100,  Angles  given  gives  ratio  of  sides  in  triangle.  101,  Right  triangle 
and  its  parts.  102,  Same  as  Art.  20.  103,  Sum  of  acute  angles  90^.  104, 
One  acute  angle  gives  form  of  right  triangle.  105,  Leg  mean  propor- 
tional between  hypothenuse  and  segment.  100,  Pythagorean.  107,  Scho- 
lium. 108,Newproof  of  Art.  100.  109,  Base  and  altitude.  110,  Rectangle 
equivalent  to  parallelogram.  Ill,  Triangle  half  rectangle.  112,  Pythago- 
rean. 113,  Criticism  invited.  —  Examples.  VIII.  Two  sides  and  included 
angle  gives  triangle.  IX.  Oblique  lines  equal.  X.  Perpendicular  line 
shortest.  XI.  Equal  sides,  equal  angles.  XII.  Equilateral,  equiangular. 
XIII.  Bisecting  isosceles  triangle.  XIV.  Equal  angles,  equal  sides. 
XV.  Equiangular,  equilateral.  XVI.  External  angle.  XVII.  Longer 
side  opposite  greater  angle.  XVIII.  Angles  in  isosceles  triangle. 
XIX.  Proof  of  Art.  93.  XX.  Square  on  diagonal.  XXI.  Quadrangle 
with  equal  opposite  angles  is  parallelogram.  XXII.  Sum  of  angles 
of  polygon.  XXIII.  Perpendicular  hypothenuse  mean  proportional. 
XXIV.  Areas  proportional  to  altitudes. 


CONTENTS.  7 

CHAPTER   VIII. 
TuE  Maximum  Area, 42 

114-116,  Circle  maximum.  117,  rolygon.  118,  Perimeter.  119, 
Isoperimetrical.  120,  Maximum.  121,  Circle  and  arc.  122,  Centre. 
12:{,  124,  Radius,  diameter.  125,  Chord.  12fl,  127,  Tangent.  128-130, 
Inscribed  and  circumacrilKid  polyg-ona,  and  circles.  131,  Regular  poly- 
gons. 132,  Analysis  of  Art,  114.  133,  Syntliesis.  134,  Area  of  triangle. 
i:]5,  Pythagorean.  136,  Shorter  line  nearer  perpendicular.  137,  Perpen- 
dicular shortest.  138,  Radius  and  tangent  at  IK)^.  139,  Radius  perpen- 
dicular to  arc.  140,  Either  side  of  triangle  shorter  than  the  sum  of  the 
others.  141,  Maximum  with  two  sides  given.  142,  Measure  of  angles. 
143,  Equal  sides  prove  equal  angles.  144,  External  angle.  145-147, 
Angle  of  two  chords,  and  corollaries.  148,  Maximum  polygon  with 
one  Hide  undetermined.  149,  Maximum  polygon  of  given  sides.  150, 
Maximum  isoperimetrical  triangle,  with  one  side  given.  151,  Perpen- 
dicular to  base  of  isosceles  triangle  bisects  the  base.  152,  New  proof 
of  Art.  150.  153,  154,  Maximum  isoperimetrical  polygon  of  given 
number  of  sides.  155,  Circle  regular  polygon.  150,  Area  of  regular 
polygon.  157,  Area  of  circle.  158,  Perimeter  of  circumscribed  polygon. 
159-101,  The  circle  the  maximum  of  isoperimetrical  figures.  162-161, 
Scholia. 


PART    II. 

CHAPTER   I. 
Geometrical  Construction,     .     . 55 

105,  Truth  invaluable.  166,  A  fortiori  valuable.  167,  Construction 
defined.  168,  Rule,  or  ruler,  how  to  make  it.  169,  Compasses.  170, 
Their  use.  171,  A  scale,  how  made.  172,  Multiplication  by  construc- 
tion.   173,  Use  of  construction. 

CHAPTER   II. 
Postulates, 59 

174,  Postulates.  175,  Plane  paper.  176,  Straight  line  from  point  to 
point.  177,  Straight  line  may  be  prolonged.  178,  Circles  may  be  drawn 
witli  any  centre  and  any  radius.    179,  Scholium. 

CHAPTER   III. 
Straight  Lines  and  Angles, 61 

180,  To  divide  a  line  mto  equal  parts^  181,  To  divide  an  angle  into 
equal  parts.  182,  Equal  chords,  equal  arcs.  183,  To  draw  angle  of  given 
degrees.  184,  Protractor.  185,  Its  use.  186,  To  draw  angle  equal  to 
given  angle.  187,  To  draw  a  line  making  given  angle  with  given  line. 
188,  Another  mode.  1S9,  Parallel  ruler,  and  triangles.  190,  191,  To 
raise  a  perpendicular  at  a  point.    192,  To  let  one  fall  from  a  point. 


8  CONTENTS. 

CHAPTER    IV. 
Triangles, 69 

193,  Three  sides  given.  194,  Two  sides  and  an  angle  given.  195,  One 
side  and  two  angles  given.  196,  Choice  of  unit!  197-199,  Impossibili- 
ties.   200-204.  Examples.    205,  200,  Scholia.    207,  208,  Peculiar  case. 

CHAPTER   V. 
Quadrangles, 72 

209,  210,  Four  sides  and  one  angle  given.  211,  Three  sides  and  two 
angles  given.    212,  Two  sides  and  three  angles  given.    213,  Examples. 

CHAPTER    VI. 
Circles, 74 

214,  To  draw  an  arc  of  given  radius.  215,  First  solution.  216,  Second 
solution.  217,  Third  solution.  218,  Fourth  solution.  219,  To  draw  a 
tangent  through  a  given  point.  220,  Examples.  221,  To  inscribe  a  cir- 
cle in  a  triangle.  222,  To  circumscribe  a  circle  about  a  triangle.  223, 
To  find  centre  of  an  arc.  224,  To  find  radius  when  the  arc  is  very 
flat.  225,  Examples.  226,  227,  To  inscribe  a  hexagon  and  equilateral 
triangle  in  a  circle.  228,  To  inscribe  a  square.  229,  To  inscribe  a  pen- 
tagon. 

CHAPTER    VII. 
Areas, 81 

230,  Area.  231,  Unit.  232,  Rectangle.  233,  Parallelogram.  234,  Tri- 
angle. 235,  Polygon.  236,  To  find  a  product.  237,  Circle.  238,  To  find  n. 
239,  240,  arcle.  241,  The  length  of  an  arc.  242,  Sector.  243,  Segment. 
244-246,  Examples. 

CHAPTER   VIII. 
Double  Position,     . 85 

247,  Approximation.  248,  Double  position.  249,  Double  position  by 
construction.    250,  Examples. 

CHAPTER   IX. 
Interpolation  and  Average, 89 

251,  Problem  explained.  252,  Application.  253,  Extension  of  double 
position.    254-256,  Examples.    257,  Scholium. 

CHAPTER   X. 
Surveying, 93 

258.  Boy's  instruments.  259,  Ten-foot  pole.  260,  Horizontal  circle. 
261,  262,  Artificial  horizon.  263,  264,  Quadrant.  265-267,  Use  of  these 
instruments.  268,  Lengths  and  angles.  269,  Lines  and  instruments  to 
be  level. 


CONTENTS.  .9 

CHAPTER   XI. 
Heights  and  Distances, 97 

270,  Hcijfht  of  tower  when  accessible.  271,  Examples.  272,  Object 
on  a  plain  seen  from  a  height.  273,  Inaccessible  tower.  274, 276,  When 
tlie  ground  is  not  level.    27C,  Example.    277-279,  Examples. 

CHAPTER   XII. 
Miscellaneous  Examples, 100 

280,  Diagonals  of  rectangle.  281,  Parallels  equidistant.  282,  One 
Bide,  one  jungle,  and  the  ratio  of  %he  pthqr  two  given.  283,  One  side  and 
the  ratio  of  tiie  three  angles  given.  284,  Circumscribed  and  inscribed 
squares.  285-288,  Angle  of  two  cliords,  and  chord  and  tangent.  289, 
Chord  and  radius  perpendicular.  290,  Two  radii  and  length  of  common 
chord  given.  291,  292,  If  two  opposite  angles  are  supplements,  the 
quadrangle  may  be  circumscribed.  293,  Chords  systematically  arranged. 
294,  Parallels  intercepted  between  parallels.  295,  Angle  between  tan- 
gents. 290, 297,  Two  sides  of  a  triangle  and  one  perpendicular  given. 
298-300,  One  wde  of  two  perpendiculars  given.  301-304,  One  angle,  with 
perpendiculars,  or  a  side,  given.  305,  To  describe  a  segment  capable  of 
containing  a  given  angle.  300,  Base,  altitude,  and  vertical  angle  given. 
307,  Case  of  equivalent  triangles,  308,  Diagonals  and  their  angle,  with 
diameter  of  circumscribed  circle,  given.  309,  Given  angle  at  the  vertex 
and  segments  of  base.  310,  311,  Diagonal  of  parallelopiped.  312, 
Square  root  of  sum  of  squares.  313,  314,  Base,  vertical  angle,  and  line 
to  the  middle  of  base,  given.  315,  310,  An  angle,  the  altitude,  and  radius 
of  circumscribed  circle  given.  iil7.  Circles  tangent.  318,  319,  Mean 
proportional  between  two  lines.  320,  321,  Tangent  and  secant.  322, 
Tripod  always  steady.  323,  Intersection  of  two  planes.  324,  Sections 
of  a  sphere  by  a  plane.    325,  32C,  Map  and  profile  of  a  railroad. 


PART    III. 

SOLID    GEOMETRY. 

CHAPTER  I. 
Ratio  and  Proportion,    . .107 

327-340,  Explanation  of  algebraic  language.  341,  Doctrine  of  propor- 
tions. 

CHAPTER   II. 
Planes  AND  Angles, .     .  110 

342-345,  Planes  and  angles.  346,  347,  Parallel  lines  in  a  plane.  348- 
351,  Perpendiculars  to  plane.  352,  Intersection  of  planes.  353,  354. 
Perpendicular  planes.     355-357,   Diedral   angle.     358,   359,   Line   and 


10  CONTENTS. 

plane.  SCO,  362,  364,  Parallel  planes.  363,  Kight  lines  not  in  a  plane. 
305,  367,  308,  Triedral  angle.  369,  When  equal.  371,  372,  Sum  of  plane 
angles 

CHAPTER    III. 

POLYEDRONS, 115 

373,  375,  Tetraedons.  376,  378,  Similar.  379-381,  Pyramids.  382, 
Prism.  383,  38t,  Parallelopiped.  385,  386,  Frustum.  387,  389,  All 
Bolids  divisible  into  pyramids. 

CHAPTER    IV. 
Areas, 119 

390-393,  Similarity.  394,  Areas  of  triangles  as  squares  on  homolo- 
gous sides.  395,  Of  polygons  the  same.  396,  397,  Surfaces  of  polye- 
drons.    398,  Pyramids  cut  by  a  plane  equidistant  from  vertices. 

CHAPTER   V. 
Volumes, 121 

399,  Unit.  400-403,  Parallelopipedon.  404-406,  Prism.  407-410,  Pyra- 
mid.   411, 412,  Truncated  prism.    413, 414,  Volumes  as  cubes. 

CHAPTER   VI. 
The  Cone, 124 

415-419,  Definitions.  420,  Volume.  422,  425,  Right  cone.  426,  Cones 
of  equal  height  as  their  bases.  427, 430,  Cylinder.  431,  Cone  one  third 
a  cylinder. 

CHAPTER   VII. 
The   Sphere, 126 

432-435,  Definitions.  436,  Radii  equal.  437,  438,  Sections  by  planes. 
439-443,  Spherical  triangle.  444,  Shortest  patli  on  a  sphere.  445, 
448,  Sum  of  sides  and  angles  in  triangle.  449-451,  Poles  and  radii. 
452,  Tangent  plane.  453-455,  Polar  triangles.  450,  457,  Three  sides 
given  gives  the  angles  of  a  spherical  triangle.  458,  Symmetrical 
triangles.  459,  The  three  angles  given  gives  the  sides  of  a  spherical 
triangle.  460,  Spherical  triangles  given  in  other  ways.  461,  Degree  of 
surface.  462,  463,  Lune.  464,  Symmetrical  triangles  equivalent.  465, 
Surface  of  triangle.  466,  467,  Surface  of  conic  frustum.  468-470,  Sur- 
face of  sphere.    471,  Solidity.    472,  Usual  decimals, 

CHAPTER   VIII. 
Problems  and  Theorems, 134 


SECOND  BOOK  IN  GEOMETRY. 


PART  I. 
CHAPTER    I. 

PRELIMINARY. 

1.  Geometry  is  the  science  of  forai.  We  really  begin 
to  learn  Geometry  when  we  first  begin  to  notice  the  forms 
of  things  about  us.  Some  pei*sons  observe  forms  much 
more  closely  than  others  do ;  partly  owing  to  their  nat- 
ural taste,  and  partly  to  their  peculiar  education.  The 
study  of  plants,  animals,  and  minerals,  the  practice  of 
drawing,  and  the  use  of  building  blocks  and  geometrical 
puzzles,  are  good  modes  of  leading  one  to  notice,  quickly 
and  accurately,  differences  of  form. 

2.  The  second  step  in  learning  Geometry  is  to  become 
able  to  imagine  perfect  forms,  without  seeing  them  drawn. 
The  Httle  book  called  "First  Lessons  in  Geometry"  was 
chiefly  designed  to  help  in  the  attainment  of  this  power. 
It  is  filled  with  descriptions  of  forms  that  cannot  be  ex- 
actly drawn.  This  is  especially  true  of  many  of  tlie 
curves,  which  cannot  be  drawn  so  exactly  as  straight-lined 
figures  and  circles,  but  which  we  can,  with  equal  ease, 
imagine  perfect. 

(11) 


12  '. ;  :  :.•'.  ;  :  definitions. 

:  -^l  .'Tiift' 't£1^4  sfep^*in  Ibaming  Geometry  is  to  learn  to 
reason  aborrt.  'foriiisi  and  to  prove  the  truth  of  the  inter- 
esting facts  which  we  think  that  we  have  observed.  This 
is  the  only  way  in  which  we  can  become  able  to  find  out 
new  truths,  and  to  be  certain  that  they  are  true.  And 
the  firet  part  of  this  second  book  is  written  to  teach  the 
scholar  how  to  reason  out,  or  prove,  geometrical  truths. 

4.  After  learning  to  reason  out  or  prove  geometrical 
tmths,  it  is  pleasant  to  know  how  to  use  them.  This  is 
not  the  only  object  of  Geometry.  It  is  worth  while  to 
know  a  truth,  simply  because  it  is  true.  But  it  is  also 
pleasant  to  be  able  to  apply  that  truth  to  practical  use,  for 
the  benefit  of  our  fellow-men.  And  the  second  part  of 
this  book  is  written  to  show  in  what  way  we  can  turn 
Geometry  to  practical  use. 


CHAPTER   II. 

DEFINITIONS. 

5.  Geometry  is  the  science  o^  form.  Every  form  or 
shape  is,  in  general,  enclosed  by  a  surface ;  every  surface 
can  be  imagined  as  bounded,  or  else  as  divided,  by  lines ; 
and  in  every  line  we  can  imagine  an  endless  number  of 
points. 

6.  A  point  is  a  place  without  any  size.  It  has  a  position, 
but  no  dimensions  ;  neither  length,  breadth,  nor  depth. 

7.  A  line  is  a  place  having  length,  without  breadth  or 
depth.  As  we  attempt  to  mark  the  position  of  a  point  by 
making  a  dot  with  the  point  of  a  pen  or  pencil,  and  the 
position  of  a  line  by  moving  the  pencil  point  along  the 
surface  of  tlie  paper,  we  find  it  convenient  to  speak  of  a 
geometrical  line  as  if  it  were  made  by  the  motion  of  a  geo- 
fnetrical  point.    As  the  eye  runs  along  the  pencil  line,  so 


DEFINITIONS.  13 

the  eye  of  the  mind  runs  along  the  geometrical  line  from 
point  to  point. 

8.  A  surface  is  a  place  having  length  and  breadth,  with- 
out depth,  that  is,  without  thickness. 

9.  A  solid  is  a  place  having  length,  breadth,  and  depth, 
A  geometrical  solid  is  not  a  solid  body,  but  is  simply  the 
space  that  a  solid  body  would  occupy,  if  it  were  of  tliat 
shape  and  in  that  place.  In  like  manner  a  geometrical 
surface  is  not  the  surface  of  a  solid  body,  but  simply  the 
surface  of  a  geometrical  solid. 

10.  A  straight  line  is  a  line  that  does  not  bend     j^ 
in  any  part.    A  point  moving  in  it  never  changes 
the  direction  of  its  motion,  unless  it  reverses  its  di- 
rection. 

11.  A  cui*ve  is  a  line  that  bends  imperceptibly  at 
every  point.     It  must  not  have  any 

^q\  straight  portion,  nor  any  corners; 
A  ^^.JV  that  is  to   say,   it  must  bend   at 

every  point,  but  the  bend  must  be 
too  small,  at  each  place,  to  be  measurable. 

12.  A  plane  is  a  geometrical  surface,  such  that    B 

a  point,  moving  in  a  straight  line  from  any  one  point  in 
the  surface  to  any  other,  never  leaves  the  surfece.  The 
common  name  of  a  plane  is  ^'ajlat  surface." 

13.  An  angle  is  the  difference  of  two  directions  in  ono 
plane.  If  the  line  C  O  should 
turn  around  the  point  O  so  as  to 
make  the  arc  D  C  grow  longer, 
the  difference  of  the  directions 
of  O  C  and  O  D  would  increase, 
and  we  should  say  that  the  angle  DOC  grew  larger  and 
larger  until  the  point  C  arrived  at  A,  so  that  the  two  lines 
O  D  and  O  C  were  opposite  in  direction. 

14.  If  the  point  C  were  earned  round  halfway  to  the 
opposite  point  A,  that  is,  to  the  point  E,  the  angle  DOC 

2 


14 


DEFINITIONS. 


-± 


would  be  a  right  angle,  as  D  O  E  is.  A  right  angle  is  a 
difference  of  direction  half  as  great  as  oppositeness  of  di- 
rection. The  difference  between  an  angle  and  a  right  an- 
gle is  called  the  complement  of  the  angle.  The  difference 
between  an  angle  and  two  right  angles  is  called  the  sup- 
plement of  the  angle.  Thus  C  O  E  is  the  complement  of 
DOC,  and  C  O  A  is  the  supplement  of  D  O  C. 

15.  When  two  lines 

make  no  angle  with  each  y/ 

other,  or  make  two  right 
angles,  they  are  called  C  — 
parallel  lines.  That  is  to  — 
say,  parallel  lines  are 
those  that  lie  in  the  same  direction  or  in  opposite  direc- 
tions. When  two  lines  in  a  plane  are  not  parallel,  the 
point  where  they  cross,  or  would  cross  if  prolonged,  is 
called  the  vertex  of  the  angle.  Lines  making  a  right  angle 
with  each  other  are  called  perpendicular  to  each  other. 

16.  A  triangle  is  a  figure  enclosed  by  three 
straight  lines  in  one  plane. 

17.  A  right  triangle  is  a  triangle  in  which  two 
of  the  sides  make  a  right  angle 
with  each  other.  These  sides 
are  then  called  the  legs  of  the 
triangle,  while  the  third  side  is 
called  the  hypothenuse. 

18.  A  parallelogram  is  a  figure 
bounded  by  four  straight  lines  in 
a  plane,  with  its  opposite  sides 
parallel. 

19.  A  rectangle  is  a  parallelo-  ^ 
gram  with  its  angles  all  right  angles. 

20.  A  square  is  a 
rectangle  with  its 
sides  all  equal. 


"/ 


REASONING.  15 

CHAPTER    III. 

REASONING. 

21.  Suppose  that  we  wished  to  make  another  person 
believe  that  the  tliree  angles  of  a 
triangle  are,  together,  equal  to  two 
right  angles.  One  way  of  convin- 
cing him  would  be  to  take  a  trian- 
gular piece  of  card,  or  of  paper,  cut 
off  the  corners  by  a  waving  line,  and  lay  the  three  comei^s 
together,  to  show  him  that  the  outer  edges  will  make  a 
straight  line,  as  two  square  comers  put  together  will  do. 

22.  Yet  he  might  not  be  satisfied  that  the  line  was  per- 
fectly straight.  Or  perhaps  he  might  say  that  if  the  angles 
of  the  triangle  were  in  a  different  proportion,  the  corners 
put  together  would  not  make  a  straight  line  with  their 
outer  edges. 

23.  A  gentleman  once  came  to  me  and  said,  "  I  have 
found  out  that  if  you  draw  such  and  such 
lines,  you  will  always  find  these  two,  A  B 
and  C  D,  equal.  At  least  my  most  careful 
measurement  shows  no  difference  between 
them ."  I  said  to  another  gentleman,  who 
knew  something  of  Geometry,  "  Can  you  prove  that  these 
lines  will  be  equal  if  the  figure  is  drawn  exactly  as  direct- 
ed ? "  He  said  he  would  try,  and  in  a  few  days  he  sent 
me  what  he  called  a  proof.  But  on  reading  it  I  found  it 
only  amounted  to  saying  that  "  if  the  lines  are  equal,  they 
are  equal."  I  then  examined  the  matter  myself,  and  found 
that  the  lines  were,  in  reality,  never  equal,  although  the 
difference  was  always  very  small,  —  too  small  to  be  easily 
discovered  by  measurement. 


C     D 


16  REASONING. 

24.  Such  errors,  too  small  to  be  discovered  by  measure- 
ment, are  sometimes  large  enough  to  do  great  mischief; 
and  at  any  rate,  however  small,  they  are  still  errors,  and 
it  is  best  to  get  rid  of  errors,  and  to  find  the  exact  truth, 
whether  the  error  is  mischievous  or  not.  In  order  to  do 
this  we  must  leani  how  to  reason,  how  to  prove  truths. 
And  in  order  to  avoid  such  mistakes  as  that  of  my  friend, 
who  thought  he  had  proved  the  false  proposition  of  which 
I  have  been  speaking,  we  must  learn  to  reason  correctly. 

25.  When  we  put  the  comers  of  a  paper  triangle  to- 
gether to  make  a  straight  line,  we  may  say,  Perhaps  there 
is  some  slight  error  here,  too  small  to  be  detected  by  meas- 
urement. How  then  shall  we  prove  that  there  is  no  such 
error  in  a  perfect  geometrical  triangle  ? 

26.  The  first  thought  that  occurs  to  us  will  be,  that  if 
any  straight  line  be  dra^vn  through  one  vertex  of  a  trian- 
gle, as  D  E  is  drawn  through  the  point  ^ 
A,  without  passing  through  the  trian-  A^ 
gle,  the  three  angles  on  one  side  of  the  -pv 
line,  about  the  point  A,  are  equal  to  two 
right  angles,  and  if  the  sum  of  the  three  C  li 
angles  of  the  triangle  is  equal  to  two  right  angles,  it  must 
be  equal  to  that  of  the  three  about  the  point  A. 

27.  But  as  the  central  angle  at  A  is  already  an  angle 
of  the  triangle,  it  follows  that  the  other  two  angles  must 
be  equal  in  their  sum  to  the  sum  of  the  angles  B  and  C 

28.  Now,  this  w^ill  be  true  in  whatever  direction  the 
line  D  E  is  drawn,  only  provided  it  does 
not  pass  through  the  triangle.  Let  us 
then  imagine  it  to  pass  in  such  a  direc- 
tion as  to  make  the  angle  B  A  E  equal 
to  the  angle  ABC,  and  it  will  only  be 

necessary  to  prove  that  D  A  C  is  then  equal  to  A  C  B. 
For  if  D  A  C  is  equal  to  A  C  B,  then,  since  we  have  sup- 
posed B  A  E  equal  to  A  B  C,  and  BAG  is  one  of  the 


REASONING.  17 

angles  of  the  triangle,  we  shall  have  the  three  angles  about 
A  equal  to  the  three  angles  of  the  triangle ;  and  as  the  three 
ano-les  about  A  are  equal  in  their  sum  to  two  right  angles, 
the  three  angles  of  the  triangle  will  be  equal  to  two  right 
angles,  which  is  what  we  wish  to  prove. 

29.  But  to  say  that  D  A  C  is  equal  to  A  C  B  is  equiva- 
lent to  saying  that  A  D  and  C  B  differ  equally  in  their 
direction  from  the  direction  of  A  C ;  and  since  A  C  is  a 
straight  line  and  its  direction  from  A  is  opposite  to  its 
direction  from  C,  this  is  equivalent  to  saying  that  AD 
and  C  B  go  in  opposite  directions. 

30.  All  that  we  have  now  to  prove  is,  that  the  line  A  D 
or  E  D  goes  in  the  same  direction  as  the  line  B  C.  But 
this  needs  no  proof,  because  we  have  already  supposed 
that  E  A  makes  the  same  angle  with  A  B  that  B  C  does ; 
and  as  A  B  is  a  straight  line,  the  direction  of  E  A  and 
C  B  must  be  opposite.  But  as  E  A  is  part  of  the  same 
straight  line  with  A  D,  it  has  the  same  direction  as  A  D. 
The  proof  is  now  complete. 

31.  And  this  mode  of  proof  does  not  depend  at  all  upon 
the  particular  shape  of  the  triangle.  We  have  made  no 
supposition  concerning  the  shape  of  A  B  C,  except  that 
it  should  be  a  triangle.  We  have,  therefore,  proved  that 
the  sum  of  the  three  angles  of  any  triangle  is  equivalent 
to  two  right  angles. 

32.  Thus  we  have  analyzed  the  proposition  that  the 
sum  of  the  three  angles  of  a  triangle  is  equivalent-  to  two 
right  angles,  and  found  that  it  resolved  itself  at  last  into 
saying  that  two  lines  making  equal  angles  on  opposite 
sides  at  the  end  of  a  straight  line  must  point  in  opposite 
directions  —  a  proposition  which  is  easily  shown  to  be 
true. 

33.  But  this  mode  of  analyzing  is  very  tedious  when 
stated  in  words.  A  geometer  usually  does  not  state  it ;  he 
passes  through  it  very  rapidly  in  his  own  mind,  and  then 

2* 


18  REASONING. 

restates  the  process  carefully  in  an  inverted  order,  as  fol- 
lows in  articles  34,  35,  and  36. 

34.  When  one  straight  line  crosses  another,  the  oppo- 
site or  vertical  angles  are  equal.  For  since  each  line  has 
but  one  direction,  the  difference  of  direction  on  one  side 
of  the  vertex  must  be  the  same  as  on  the  other  side. 

35.  When  one  straight  line  crosses  two  parallel  straight 
lines,  the  alternate  inter- 

nal  angles  are  equal,  or  j^ 

in  the  figure   A  G  E   is  ~7g  ^ 

equal   to    D  H  F.      For    c "Zl D 


DHF  is  equal  to  B  GF,        y^ 

having  its  sides  pointing     '  ^ 

-in  the  same  direction  as  those  of  B  G  F,  and  A  G  E  is 

equal  to  B  G  F  by  article  34. 

36.  Through  the  vertex  of  any  triangle,  as  through  A, 
draw  a  straight  line  D  E  parallel  to  the 

opposite  side  B  C.     Then  E  A  B  will  be  D -tJt ^ 

equal  to    its    alternate   internal    angle         y^  \  \ 

ABC;  and  for  the  same  reason  D  A  C    c  -^ ^ — -^B 

will  be   equal  to  A  C  B.     So  that  the 
three  angles  of  the  triangle  will  be  equal  to  the  three  an- 
gles about  the  point  A,  and  their  sum  is  plainly  two  right 
angles. 

37.  The  mode  of  proving  that  the  sum  of  the  three  an- 
gles of  a  triangle  is  equal  to  two  right  angles,  by  cutting 
a  piece  of  card,  is  called  experimental  proof.  It  is  of  very 
little  use  in  mathematics,  but  of  great  use  in  the  study  of 
physics,  especially  in  mechanics  and  chemistry. 

38.  The  mode  of  proof  used  in  articles  26-31  is  called, 
by  metaphysicians  and  by  writers  on  Arithmetic,  analysis. 
But  as  geometers,  in  their  writings,  almost  never  use  this 
method,  they  have  no  name  for  it;  and  when  they  speak 
of  analysis,  or  of  analytical  methods,  they  usually  refer  to 
something  else  apparently  of  a  very  different  <;liaracter. 


ANALYSIS  AND  SYNTHESIS.  19 

39.  The  mode  of  proof  in  articles  34-36,  called  by 
metaphysicians  synthesis,  by  geometers  demonstration 
or  deduction,  is  that  usually  employed  in  stating  geomet- 
rical results.  This  mode  is  chiefly  applicable  to  mathe- 
matics, and  must  be  used  with  very  great  caution  in  rea- 
soning upon  other  subjects. 

40.  A  proposition,  which  we  wish  to  prove,  may  be 
compared  to  a  mountain  peak  which  we  wish  to  show  is 
accessible  from  the  highway.  The  method  of  articles 
26-31  may  be  compared  to  taking  a  flight  by  a  balloon  to 
the  top  of  the  peak,  and  then  finding  a  path  down  to  the 
highway;  while  the  method  of  articles  34-36  maybe  com- 
pared to  the  direct  ascent  of  the  mountain.  In  either  case 
we  show  that  the  peak  is  accessible,  because  we  actually 
pass  over  all  the  steps  of  a  connected  pathway  between 
the  road  and  the  mountain  top. 

Thus  in  geometrical  demonstration  we  pass  through 
every  step  connecting  the  simplest  self-evident  truths  with 
the  highest  deductions  of  the  science  ;  while  in  the  process 
which  writers  on  Arithmetic  call  analysis,  we  pass  over 
ever^  step  from  the  latter  truths  down  to  the  simplest. 
In  either  case  we  prove  that  the  higher  truth  really  stands 
on  the  same  basis  as  the  simpler,  and  must,  therefore,  be 
true. 


CHAPTER    IV. 

ANALYSIS    AND    SYNTHESIS. 

41.  What  I  have  called  demonstration  or  deduction, 
but  which  is  better  called  synthesis,  because  it  is  a  putting 
together,  one  by  one,  of  the  parts  of  a  complex  truth,  is  the 
only  mode  of  proof  that  you  will  usually  find  in  works  on 


20  ANALYSIS    AND    SYNTHESIS. 

geometry.  And  if  such  works  are  carefully  read  they  are 
always  intelligible  to  a  child  of  good  geometrical  reason- 
ing powers. 

42.  But  the  study  of  such  works  does  not  always  teach 
a  child  to  reason  for  himself.  The  pupil  says,  "  Yes,  I  un- 
derstand all  this,  and  yet  I  could  not  have  done  it  without 
aid ;  I  do  not  see  how  the  writer  knew  where  to  begin ; 
how  he  knew  that  by  starting  from  these  particular  truths, 
and  going  in  that  particular  path,  he  could  reach  that 
proposition."  A  pupil  who  had  never  studied  geometry 
could  not,  for  instance,  tell  why  in  articles  34-36  we  should 
begin  with  showing  that  vertical  angles  are  equal.  He 
would  not  see  any  contiection  between  that  truth  and  the 
desired  proof,  and  would  not  know  that  this  synthesis  had 
been  preceded,  in  the  mind  of  the  writer,  by  a  rapid  anal- 
ysis, such  as  that  of  articles  26-31. 

43.  It  is  as  though  a  mountain  guide,  wishing  to  make 
for  a  child  a  path  up  to  a  mountain  peak,  should  lead  him 
along  the  highway  until  the  peak  was  hidden  by  the 
lower  hills  about  its  base,  and  then  begin  boldly  to  clear 
a  road,  through  the  brush-wood  and  trees,  until  he  reached 
the  top.  The  child  might  say.  "  How  did  you  dare  begin 
at  once  to  cut  down  the  bushes  and  clear  the  path  ?  How 
did  you  know  that  the  road  you  were  making  would  not 
lead  you  to  the  edge,  or  to  the  foot,  of  some  precipice,  or 
that  it  would  not  take  you  to  a  different  peak  from  that 
which  you  wished  to  climb  ?"  And  if  the  child  received 
no  answer  to  his  questions,  —  if  he  was  not  told  that  the 
guide  had  already  climbed  to  the  summit  and  again  de- 
scended,—  he  would  have  learned  little  to  help  him  in 
laying  out  paths  for  himself. 

44.  In  like  manner,  although  the  descent  from  difficult 
propositions  to  more  simple  is  more  tedious  than  the  as- 
cent, it  will  be  more  useful  to  a  learner,  because  it  will 
show  him  the  maimer  in  which,  by  a  mental  process,  we 


ANALYSIS   AND    SYNTHESIS.  21 

discover  the  points  from  wliich  wc  are  to-  start  in  our 
ascent.  That  is  to  say,  if  we  follow  a  good  analysis,  we 
shall  learn  how  to  perform  synthesis  for  ourselves ;  but  if 
we  were  simply  to  follow  a  writer's  synthesis,  we  should 
not  learn  how  to  analyze,  which  must  nevertheless  always 
go  before  synthesis, 

45.  Among  the  first  requisites  in  reasoning  is  a  clear 
understanding  of  the  object  in  view  ;  that  is,  of  the  point 
to  be  proved ;  and  next,  a  clear  perception  of  each  partic- 
ular part  of  the  demonstration,  and  of  the  connection  of 
each  part  wuth  the  adjacent  parts. 

Thus,  in  laying  out  a  path  up  a  mountain,  it  is  necessary 
to  know  exactly  from  what  point  yon  wish  to  start,  and  to 
what  point  you  wish  to  go.  It  is  also  necessary  to  exam- 
ine carefully  each  point  of  the  road,  for  a  single  impassable 
place  would  destroy  the  value  of  the  whole  road. 

46.  Each  step  of  the  proof  must  be  a  simple  step,  and 
clearly  true ;  that  is,  it  must  be  so  simple  and  self-evident 
as  to  be  beyond  all  doubt. 

47.  The  analysis  must  end,  or  the  synthesis  begin,  with 
truths  that  are  self-evident,  or  else  that  have  been  already 
proved.  Your  mountain  path  must  begin  on  level,  or  at 
least  on  accessible  ground. 

48.  Care  must  be  taken  not  to  introduce  any  thing  as 
true  which  has  not  been  proved.  This  would  be  like 
starting  your  mountain  road  in  two  places  at  once.  You 
might  afterwards  find  impassable  barriei-s  between  the  two 
parts  of  your  road,  and  perhaps  find  that  one  of  them  could 
not  be  made  to  the  top  of  the  mountain,  nor  the  other  to 
its  base.  For  example,  in  Art.  36,  I  drew  a  straight  line 
through  A,  parallel  to  B  C.  This  was  very  well  —  for  no 
one  can  possibly  doubt  such  a  line  might  be  drawn.  But 
if,  instead  of  that,  I  had  said.  Let  us  draw  a  straight  line 
through  A  in  such  a  manner  as  to  make  the  angles  on 
the  two  sides  of  A  equal  to  the  angles  B  and  C,  I  should 


22  ANALYSIS    AND    SYNTHESIS. 

have  done  what  I  had  no  right  to  do.  For  that  would  be 
taking  for  granted  a  thing  which  I  must  prove ;  namely, 
that  a  straight  line  can  be  thus  drawn.  It  would  be  start- 
ing half  way  up  my  mountain,  and  taking  for  granted  that 
the  lower  part  of  the  path  could  be  built  afterwards.  It 
would  require  a  straight  line  to  fulfil  two  conditions  at 
once,  without  having  shown  that  one  condition  does  not 
exclude  the  other. 

49.  Whether  we  reason  by  synthesis  or  analysis,  we 
must  therefore  reason  very  carefully,  in  order  to  connect 
the  proposition  which  we  wish  to  prove  by  a  stairway  of 
self-evident  steps  with  a  self-evident  foundation. 

50.  By  a  self-evident  truth,  I  mean  a  truth  which  can- 
not be  made  any  plainer,  and  which  is  already  perfectly 
plain  to  an  intelligent  person  who  looks  steadily  at  it. 
For  instance,  that  two  straight  lines  can  cross  each  other 
only  in  one  place  at  once ;  that  any  cui-ve  can  be  cut  by  a 
straight  line  in  at  least  two  places ;  that  either  side  of  a 
triangle  is  shorter  than  the  sum  of  the  other  two ;  tl\at  if 
three  strings,  and  no  more  than  three,  come  from  one 
point,  one  of  them  must  have  an  end  at  that  point,  — 
these  are  self-evident  truths. 

51.  By  a  self-evident  step  in  reasoning,  I  mean  the 
statement  of  the  relation  of  one  truth  to  another,  or  of  the 
dependence  of  one  truth  upon  another,  when  that  depend- 
ence or  that  relation  is  itself  a  self-evident  truth.  Self 
evident  steps  in  reasoning  are  simply  the  statement  of 
self-evident  truths  of  connection.  For  instance,  when  w^ 
have  explained  the  meaning  of  "  a  straight  line  "  by  calling 
it  a  line  that  has  in  every  part  the  same  direction,  and 
have  explainedthe  word  "angle"  to  mean  the  difference  of 
two  directions  in  one  plane,  then  it  follows  that  the  angle 
which  two  straight  lines  make  with  each  other  is  the  same 
in  one  part  of  the  lines  as  in  any  other,  and  that  tlie  two 
different  angles  apparently  made  by  two  straight  lines 


ANALYSIS  AND  SYNTHESIS.  23 

cannot  really  bo  made,  unless  one  of  the  lines  goes  in  two 
o[)posite  directions  at  the  same  time.  No  reasoning  can 
make  the  connection  between  these  definitions  and  the 
equality  of  vertical  angles  any  more  plain.  It  is  a  self- 
evident  connection. 

52.  Or,  suppose  that  we  say  that  you  cannot  make  one 
rope  go  from  a  centre  post  to  the  four  corners  of  a  square, 
and  also  around  the  square,  and  have  but  a  single  rope 
from  post  to  post.  We  should  prove  it  in  this 
way.  Let  there  be  a  rope  around  the  square, 
and  going  also  from  each  post  to  the  centre. 
This  of  course  can  be  imagined.  It  is  a  defi- 
nite and  allowable  conception.  But  we  will  prove  that 
this  rope  must  be  in  two  pieces.  For  each  of  the  four 
corners  will  have  three  lines  coming  from  it,  one  towards 
each  adjacent  corner,  and  one  towards  the  centre.  Thus 
it  follows  by  self-evident  connection,  from  the  conception 
of  the  rope  going  around  the  square  and  to  each  corner, 
that  there  will  be  four  points,  from  each  of  which  three 
lines  come.  But  it  is  a  self-evident  truth  that  at  each  of 
these  points  there  must  be  one  end  of  a  rope.  Hence,  by 
self-evident  connection,  there  will  be  four  ends  of  rope 
about  the  square.  Hence,  by  self-evident  connection  with 
the  self-evident  truth  that  one  piece  of  rope  can  have  but 
two,  and  must  have  two,  ends,  it  follows  that  there  must 
be  two  pieces  of  rope,  and  cannot  be  only  one.  Now,  the 
whole  of  this  proof  is  simply  the  statement  of  self-evident 
connections  betvv^een  the  proposition  that  one  rope  cannot 
go  around  a  square  and  also  from  each  corner  to  the  cen- 
tre without  doubling,  and  the  self-evident  truths  that  a 
piece  of  rope  must  have  two,  and  cannot  have  more  than 
two,  ends ;  and  that  when  only  three  lines  of  rope  come 
from  one  point,  one  of  them  must  end  at  that  point.  The 
proof  is  simple  ;  and  yet  intelligent  men  have  spent  hours 
in  experimenting  with  a  string  and  five  posts  thus  arranged, 
or  with  a  pencil  and  five  dots  representing  posts. 


24  VARIETY    OP   PATHS. 

53.  Many  self-evident  truths  are  general,  and  self-evi- 
dent steps  are  generally  the  recognition  of  general  rela- 
tions; and  therefore  most  writers  on  reasoning  say  that 
reasoning  consists  simply  in  showing  that  a  particular  case 
comes  under  a  general  class,  that  is,  that  the  only  self- 
evident  connection  of  propositions  is  the  actual  inclusion  of 
one  proposition  in  another.  But  in  the  mathematics,  there 
are  many  self-evident  truths  which  it  is  difficult  to  state  in 
a  general  form ;  and  I  therefore  think  that  the  explanation 
which  I  have  given  of  the  process  of  reasoning  will  be  of 
more  use  to  you  in  your  geometrical  studies. 


CHAPTER   V. 

VARIETY   or   PATHS. 

54.  As  there  are  usually  many  paths  by  which  we  may 
ascend  a  hill,  so  there  are  usually  many  modes  by  which 
we  may  demonstrate  a  proposition.  In  the  case  of  a  sim- 
ple proposition,  it  is  not  usually  worth  while  to  try  more 
than  one  mode.  But  with  more  difficult  problems,  it  is 
sometimes  worth  while  to  spend  a  great  deal  of  labor  in 
discovering  the  simplest  mode  of  demonstration.  There 
are  geometrical  truths  which  can  be  demonstrated  in  so 
simple  a  manner  as  to  require  only  twenty  lines  to  write 
down  the  demonstration ;  and  yet  some  writers,  from  ig- 
norance of  this  simple  mode,  have  written  more  thantwent)^ 
pages  to  prove  the  same  truths. 

55.  It  will  therefore  be  useful  to  you,  to  show  you,  by 
a  simple  example,  such  as  that  of  the  equality  of  the  sum 
of  the  angles  in  a  triangle  to  two  right  angles,  the  great 
variety  of  methods  by  which  a  single  proposition  can  be 
proved. 


VARIETY   OF   PATHS. 


25 


56.  In  the  proofs  of  this  proposition,  which  I  will  now 
give  yon,  I  will  not  be  careful  to  follow  out  every  step. 
It  will  be  enough, /br  the  purpose  ice  now  have  in  vieWy 
simply  to  show  you  the  general  line  of  the  paths,  without 
taking  you  through  every  step  of  the  way. 

57.  The  line  D  E  might  be  drawn  so  as  to  coincide  in 
direction  with  one  of  the  other  sides 

of  the  triangle,  as  A  B,  which  would 
give  us  the  figure  in  the  margin.  And 
by  imagining  the  dotted  line  A  F  par- 
allel to  C  B,  we  should  have  F  A  C 
equal  to  A  C  B,  and  FAD  equal  to 
C  B  A,  which  would  make  the  three  angles  at  A  equal  to 
the  three  angles  of  the  triangle,  as  in  the  former  proof. 
(See  Chapter  III.) 

58.  By  prolonging  the  lines  C  A  and  B  A  through  the 
^oint  A,  D  E  being  parallel  to  B  C, 

we  should  have  NAD  equal  to   ^  \A^ 

ABC,  JAE  equal  to  A  CB,  and 

NAJ  equal  to  CAB.     So  that 

the  three  angles  of  the  triangle  will 

be  equal  to  the  three  angles  on  the  upper  side  of  the  line 

D  E,  which  are  manifestly  equal  to  two  right  angles. 

59.  The  three  methods  of  proving  this  proposition  that 
I  have  now  given,  are  strictly  geometrical.  Others  might 
be  given,  that  are  something  more  like  algebraic  reasoning. 

60.  Let  us,  for  instance,  imagine  each  side  of  a  triangle 
prolonged  at  its  right  hand  end, 
as  in  this  figure,  and  also  a  line  ^ 
drawn  from  one  vertex,  as  C,  \ 
parallel  to  the  opposite  side  B  A. 
Now,  the  external  angles  F  C  B,  ^ 
D  B  A,  E  A  F,  are  plainly  equal  ^ 
to  the  three  angles  F  C  B,  B  C  G,  G  C  F,  and  these  amount 
to  four  right  angles.    But  each  external  angle  is  plainly 

3 


26  VAEIETY   OF   PATHS. 

the  supplement  of  one  of  the  angles  of  the  triangle  ;  that 
is,  it  is  equal  to  the  difference  between  two  right  angles 
and  one  angle  of  the  triangle.  The  sum  of  the  three  ex- 
ternal angles  must  therefore  be  equal  to  the  difference 
between  six  right  angles  and  the  angles  of  the  triangle. 
But  as  this  difference  is  four  right  angles,  the  three  angles 
of  the  triangle  must  be  equivalent  to  two  right  angles. 

61.  If  we  introduce  the  idea  of  motion,  we  can  devise 
quite  a  different  sort  of  demonstration.  p 
Suppose,  for  instance,  that  I  stand  at  the 
point  A,  w^ith  my  face  towards  C.  Let 
me  now  turn  to  the  right  until  I  face 
towards  B.  I  have  now  changed  the 
direction  of  my  flice  by  an  amount  w^hich 
is  equal  to  the  angle  at  A.  Suj)pose  that  I  now  walk  to 
B,  without  turning ;  I  shall  have  my  back  towards  A,  and 
if  standing  still,  I  turn  to  the  right,  until  I  have  cliangcd 
my  direction  by  an  amount  equal  to  the  angle  ABC;  I 
shall  have  my  back  towards  C.  Let  me  now  walk  back- 
ward without  turning,  until  I  reach  C,  and  I  shall  have 
my  face  tOAvards  B.  I  will  now  turn  a  third  time  to  tlie 
right,  until  I  face  the  point  A.  My  three  turnhigs,  or 
changes  of  direction,  have  been  equal  to  the  three  angles 
of  the  triangle;  they  have  all  been  to  the  right ;  thci-efore 
my  whole  change  of  direction  is  equal  to  tlie  sum  of  these 
angles ;  I  am  now  looking  in  exactly  tlie  opposite  direction 
to  that  from  which  I  started  ;  I  am  looking  from  C  to  A, 
instead  of  from  A  to  C;  I  have  turned  half  way  round; 
that  is,  through  two  right  angles.  Whence,  the  sum  of 
the  three  angles  of  the  triangle  is  equivalent  to  tv/o  right 
angles. 

62.  Another  demonstration,  by  means  of  motion,  may 
be  obtained  as  follows :  Suj)posc  an  arrow,  longer  than 
either  side  of  the  triangle,  to  be  laid  upon  the  side  A  C, 
pointing  in  the  direction  from  A  to  C.  Taking  hold  of  the 
pointed  end  beyond  C,  turn  the  arrow  round  upon  the 


VARIETY    OF   PATHS.  27 

point  A,  as  a  pivot,  until  the  arrow  lies  upon  the  line  A  B. 
Taking  now  hold  of  the  further  end,  beyond  A,  turn  the 
aiTow  upon  B  as  a  pivot,  until  the  arrow  lies  upon  the 
line  B  C.  Using  C  as  a  pivot,  turn  it  now  until  the  point 
of  the  arrow  is  over  A.  The  arrow  has  thus  been  reversed 
in  direction,  turned  half  way  round,  or  through  two  right 
angles.  It  has  been  turned  successively  through  tlie  three 
angles  of  a  triangle,  and  every  time  in  the  same  direction, 
like  the  hands  of  a  watch ;  so  that  its  total  change  of 
direction,  two  right  angles,  is  equivalent  to  the  sum  of  the 
three  angles. 

63.  You  have  thus  seen  how  a  single  proposition  may 
be  proved  in  a  variety  of  ways.  We  have  shown  what  is 
the  value  of  the  sum  of  the  angles  in  a  triangle  in  six 
different  ways  ;  in  three,  by  what  is  called  rigid  geometry ; 
in  one,  by  a  partly  algebraical  process ;  and  in  two,  by 
introducing  the  idea  of  motion.  And  I  wish  you  to  ob- 
serve, that  every  one  of  the  six  ways  is  satisfactory.  They 
are  all  proofs  that  are  certain,  because  they  lead  you  from 
self-evident  truths  by  self-evident  steps.  One  is  not  more 
certain  than  the  other,  because  they  are  all  absolutely  cer- 
tain. The  only  choice  between  them  is,  that  some  are  more 
purely  geometrical ;  some  are  better  adapted  to  the  peculiar 
tastes  of  different  students ;  and  some  are  neater,  and  more 
quickly  perceived  by  untaught  persons. 

Examples. 

By  aid  of  the  principles  and  methods  of  the  five  preced- 
ing chapters,  the  learner  may  demonstrate  (sometimes  in 
a  variety  of  Avays)  the  following  simple  projiositions  X  — 

I.   When  parallel  lines 

are  crossed  by  a  third,  the /^ 

external '  internal  angles  ~7g 

are  equal ;  that  is,  F  G  B    C ^/^.  ■   ■ D 

=  GHD,4fcc.  X 


28  THE   PYTHAGOREAN   PROPOSITION. 

II.  If  two  lines,'  cut  by  a  third,  make  the  alternate- 
internal,  the  external-internal  or  the  opposite-external 
angles  equal,  the  lines  are  parallel. 

III.  If  two  lines,  crossed  by  a  third,  make  the  adjacent 
internal  angles  (as  B  G  H,  D  H  G)  supplements  to  each 
other,  the  lines  are  parallel. 

IV.  If  two  lines  make  the  same  angle  with  a  third,  they 
are  parallel  to  each  other. 

V.  State  this  proposition  for  the  cases  when  the  angle 
is  zero,  one  right  angle,  and  two  right  angles. 

VI.  Parallel  lines  can  never  meet.  [Note.  To  prove 
a  negative  of  this  kind,  the  easiest  mode  is  to  show  the 
absurdity  of  the  affirmative.  In  the  present  case,  grant 
that  the  lines  met  at  a  certain  point,  and  show  from  the 
nature  of  the  straight  line,  that  the  parallel  lines  must  in 
this  case  be  one  line,  which  is  absurd.] 

VII.  Only  one  perpendicular  can  pass  through  a  given 
point  to  a  given  straight  line.     [Proof  by  VI.] 


CHAPTER   VI. 

the    PYTHAGOREAN   PROPOSITION. 

64.  YoTJ  recollect  that  the  square  built  on  the  hypoth- 
enuse  of  a  right  triangle  is  equivalent  in  its  area  to  the 
sum  of  the  squares  built  upon  its  legs. 
This  is  one  of  the  most  useful  of  all 
geometrical  truths.  Let  us  first  an- 
alyze it  in  one  or  two  modes,  and 
then  build  it  up  synthetically  by  the 
same  paths.  We  may  afterwards,  if 
we  like,  devise  other  modes  of  anal- 
ysis and  synthesis ;  for  this  proposi- 
tion, like  all  others,  may  be  ap- 
proached in  various  ways. 


THE    PYTHAGOREAN   PROPOSITION. 


29 


65.  The  Pythagorean  proposition  or  theorem  might  bo 
suggested  in  different  ways.  But  in  whatever  way  wo 
were  led  to  suspect  that  the  square  on  the  liypothenuse 
is  equivalent  to  the  sum  of  the  squares  on  the  legs,  we 
should,  in  reflecting  upon  it,  probably  begin  by  drawing  a 
right  triangle  with  a  square  built  upon  each  side. 

CG.  We  should  inquire  whether  the  square  on  the  hy- 
pothenuse  could  be  divided  into  two  parts  that  should  be 
respectively  equal  to  the  other  two  squares.  And  we 
should  judge  that  these  parts  should  be  somewhat  similar 
to  each  otlier  in  shape,  because  the  legs  do  not  differ  in 
their  relations  to  the  hypothenuse,  except  in  size,  and  in 
the  angles  they  make  with  it. 

67.  But  we  cannot  readily  conceive  of  any  division  of 
the  square  into  two  somewhat  similar  parts,  except  into 
two  rectangles.  And  then  it  is  ap- 
parent that  two  rectangles,  bearing 
respectively  the  same  relations  to 
the  squares  on  the  legs,  may  be 
formed  by  drawing  a  line  from  the 
vertex  of  the  right  angle  at  right 
angles  with  the  hypothenuse,  and 
continuing  it  through  the  square,  as 
C  F  is  here  drawn. 

68.  It  will  now  only  be  necessary 
to  show  that  one  of  these  rectangles  is  equivalent  to  its 
corresponding  square ;  because  the  same  mode  of  proof  will 
obviously  answer  for  the  other  rectangle  and  its  square. 

69.  Now,  if  we  know,  or  can  prove,  that  the  area  of  a 
rectangle  is  measured  by  the  product  of  its  sides,  we  shall 
have  to  prove  that  A  E  X  A  B,  or  A  E  X  A  B,  is  equiva- 
lent to  A  C  X  A  e. 

70.  But  by  the  doctrine  of  proportion  it  may  be  shown 
that  this  would  be  equivalent  to  saying  that  A  E  is  to  A  C 
35  A  C  is  to  A  B. 

3* 


30  THE    rYTIIAGOREAN    niOrOSITIOX. 

71.  Again,  it  may  be  shown  by  geometry  that  this  pro- 
portion between  the  lines  A  B,  A  C,  and  A  E,  would  be 
true  if  the  triangle  A  E  C  were  similar  to  A  C  B,  and  that 
A  E  stood  in  one  to  AC  as  A  C  stood  to  A  B  in  the 
other ;  so  that  all  that  remains  for  us  to  do  is  to  show  that 
these  triangles  are  similar. 

72.  But  we  can  show  by  geometry  that  two  triangles 
are  similar  when  their  angles  are  equal. 

73.  And  it  is  easy  to  show  that  the  angles  of  these  tri- 
angles are  equal  to  each  other. 

74.  For  CAB  and  C  A  E  are  the  same  angle ;  A  C  B 
and  A  E  C  are  both  right  angles ;  and  therefore  ABC  and 
ACE  are  each  complements  of  C  A  E.  Moreover,  A  C 
and  A  E  are  situated  in  the  triangle  A  E  C,  in  the  same 
manner  that  A  B  and  A  C  are  situated  in  the  triangle 
ABC. 

75.  We  have  thus,  in  articles  66-74,  sufficiently  ana- 
lyzed the  Pythagorean  proposition  to  enable  us  to  build  it 
up  again  in  a  deductive  form.  This  analysis,  however, 
has  been  partly  algebraical,  as  it  has  introduced  the  idea 
of  multiplying  two  lines  to  produce  a  surface.  Let  us 
now  begin  and  build  up  the  proposition  by  the  same  road. 
We  shall  find  31  articles  necessary ;  and  I  will  number 
them  from  76  to  106. 

First  Proof  of  the  Pythagorean  Proposition. 

76.  Pefinition,  The  comparative  size  of  two  quantities 
is  called  their  ratio ;  thus,  if  one  is  twice  as  large  as  the 
other,  they  are  said  to  be  in  the  same  ratio  as  that  of  2  to 
1 ;  or  to  be  in  the  ratio  2  to  1 ;  or  it  is  said,  in  a  looser 
way,  that  their  ratio  equals  2. 

77.  Notation,  Ratio  is  written  by  means  of  the  marks 
:,  —,  and  by  writing  one  quantity  over  the  other.     Thus, 

A :  B,  A  -f-  B,  and  jg,  are  each  used  to  signify  the  ratio 


THE   PYTHAGOREAN   PROPOSITION.  81 

of  A  to  1^>.  These  marks  arc  the  same  as  those  used  in 
iirithni'v  lie.  to  .sigiiity  Quotient,  because  the  meanmg  of  a 
(juotiuiit  is  '-d  number  havmg  the  same  ratio  to  1  that 
the  dividend  has  to  the  divisor."  The  ratio  of  A  to  B  is 
not  the  quotient  of  A  divided  by  13,  but  it  is  the  ratio  of 
that  quotient  to  iniity. 

78.  Axiom.  If  each  of  two  quantities  is  multipUed  or 
divided  by  the  same  number,  the  ratio  of  the  products  or 
quotients  will  be  tlie  same  as  that  of  the  quantities  them- 
selves. Thus  twenty  inches  is  in  the  same  ratio  to  twenty' 
rods  as  one  inch  to  one  rod,  or  as  the  twentieth  of  an  inch 
to  the  twentieth  of  a  rod. 

79.  Dejinition.  A  proportion  is  the  equality  of  two 
r:itios.  Thus  (if  we  use  the  sign  i=  to  signify  "is  equal 
to  ")  A  :  13  =  C  :  D  is  the  statement  of  a  proportion.  It 
signilies  that  A  is  in  the  same  ratio  to  B  that  C  is  to  D. 

80.  Definition,  When  a  proportion  is  written  as  in 
article  79,  the  first  and  last  terms,  that  is,  A  and  D,  are 
called  the  extremes,  and  the  others,  that  is,  B  and  C,  arc 
c;illed  the  means. 

81.  Theorem,  In  every  proportion  the  product  of  the 
means  is  equal  to  that  of  the  extremes.  Proof,  In  any 
pro]^ortion,  as  M  :  N"  =  P  :  Q,  we  wish  to  prove  (using  the 
mark  X  to  signify  «  multipHed  by")  that  M  X  Q  =  N  X  P. 
Xow,  in  order  to  do  this,  we  must  use  only  self-evident 
truths.  The  only  truth  of  this  character  that  we  have 
given  above  is  that  of  article  78.  But  in  order,  by  means 
of  the  multiplications  of  article  78,  to  change  the  first  ratio 
]\[ :  N^  into  M  X  Q,  we  must,  whatever  else  we  do,  at  least 
m  ultiply  each  term  by  Q,  and  this  will  give  us  M  X  Q : 
X  X  Q  =  P  :  Q ;  and  in  order  to  change  the  second  ratio 
P :  Q  into  N  X  P?  we  must,  at  all  events,  multiply  each 
term  by  N,  and  this  will  give  us  M  X  Q  •  ^  X  Q  = 
N  X  P  :  N  X  Q. 

Thus,  from  the  self-evident  truth  of  article  78,  we  find 


32  THE   PYTHAGOREAN   PROPOSITION. 

that  the  product  of  the  means  bears  the  same  ratio  to  the 
product  N  X  Q  that  is  borne  to  it  by  the  product  of  the 
extremes.  And  as  it  is  self-evident  that  two  quantities, 
bearing  the  same  ratio  to  a  third,  must  be  equal  to  each 
other,  we  have  proved  that  the  product  of  the  means  is 
equal  to  that  of  the  extremes. 

82.  Definition,  When  both  the  means  are  the  'same 
quantity,  that  quantity  is  called  a  mean  proportional  be- 
tween the  extremes. 

83.  Corollary.  It  follows  from  article  81,  that  the 
product  of  the  mean  proportional  multiplied  by  itself  is 
equal  to  the  product  of  the  extremes. 

84.  Definitions »  A  unit  of  length  is  a  line  taken  as  a 
standard  of  comparison  for  lengths.  Thus  an  inch,  a  foot, 
a  pace,  a  span,  &c.,  are  units.  The  length  of  any  line  is 
its  ratio  to  the  unit  of  length. 

85.  Definition,  A  unit  of  surface  is  a  surface  taken 
as  a  standard  of  comparison.  The  most  common  unit  of 
surface  is  a  square  whose  side  is  a  unit  of  length. 

86.  Definition.  The  area  of  a  surface  is  the  ratio  of 
the  surface  to  the  unit  of  surface. 

87.  Theorem,  Any  straight  line  in  the  same  plane  with 
two  parallel  lines  makes  the  same  angle  with  one  that  it 
does  with  the  other.  Proof,  For  as  the  straight  line  has 
but  one  direction,  and  each  of  the  parallel  lines  may  al- 
ways be  considered  as  going  in  the  same  direction  as  the 
other,  the  difference  of  that  direction  from  the  direction  of 
the  third  straight  line  must  be  the  same  for  each  of  the 
parallel  lines. 

88.  Corollary,  If  a  straight  hne  is  parallel  to  one  of 
two  parallel  lines,  it  is  parallel  to  the  other ;  if  at  right 
angles  to  one  of  the  two,  it  is  at  right  angles  to  the  other. 

89.  Theorem,  If  a  straight  line  makes  on  the  same  side 
of  itself  the  same  angle  with  two  other  straight  lines  in  the 
same  plane,  those  other  straight  lines  must  be  parallel. 


THE   PYTHAGOREAN   PROPOSITION.  33 

Scholium.  The  line  must  not  be  conceived  as  reversing 
its  direction  at  any  point.  Proof,  For  if  two  directions 
differ  equally  from  a  third,  they  must  be  equal  to  each 
oUier. 

Second  Scholium,  If  the  straight  line  reverses  its  di- 
rection between  the  other  lines,  and  makes  equal  angles 
with  them,  it  shows  that  it  crosses  each  at  an  equal  dis- 
tance from  their  point  of  mutual  intersection. 

90.  Axiom,  If  the  boundaries  of  one  plane  surface  are 
similar  to  those  of  another  in  such  a  way  that  the  two 
surfaces  would  coincide  in  extent  if  laid  one  upon  the 
other,  the  two  surfaces  are  equivalent. 


CHAPTER   VII. 

THE    PYTHAGOREAN    PROPOSITION  CONTINUED 

91.  Theorem,  If  a  triangle  has  one  side  and  the  adja- 
cent angles  equal  respectively  to  a  side  and  the  adjacent 
angles  in  another  triangle,  the  two  ^  j, 

triangles  are  equal.  Proof,  Let  us 
suppose  that,  in  the  triangles  ABC 
and  D  E  F,  we  have  the  side  A  B 
equal  to  the  side  D  E,  the  angle  at 

A  equal  to  the  angle  at  D,  and  that  at  B  equal  to  that  at  E. 
Let  us  imagine  the  triangle  D  E  F  to  be  laid  upon  ABC 
in  such  a  manner  as  to  place  E  upon  B,  and  D  upon  A, 
which  can  be  done,  because  A  B  is  equal  to  D  E.  Now,  as 
the  angle  A  is  equal  to  D,  the  line  D  F  will  run  in  the 
same  direction  as  A  C,  and,  as  it  starts  from  the  same 
point,  will  coincide  with  it.  Also,  since  the  angle  B  is 
equal  to  E,  the  line  E  F  will  coincide  wnth  B  C.      The 


34  THE    PYTHAGOREAN   PROPOSITION. 

point  (F)  of  intersection  of  D  F  and  E  F  must  therefore 
coincide  witli  C,  the  point  of  intersection  of  A  C  and 
B  C.     Whence,  by  article  90,  the  triangles  are  equal. 

92.  Theorem,  The  opposite  sides  of  a  parallelogram 
are  equal.  Proof,  Article  90  gives  us  the  only  test  of 
geometrical  equality.  So  that,  in  order  to  prove  this  the- 
orem, we  must  show  that  in  a  parallelogram  like  A  B  C  D, 
A  B  may  be  made  to 

coincide  with  D  C,  5^— . -jj^yC 

and  B  C  with  A  D. 
And  this  would  ev- 
idently be  done  if 
we  could  show  that 
the  triangle  A  B  C  is  equal  to  A  D  C.  But  in  these  tri- 
angles the  line  A  C  is  the  same,,  and  by  article  87  the  adja- 
cent angles  A  C  B  and  CAB  are  equal  to  the  adjacent 
angles  CAD  and  A  C  D ;  whence,  by  article  91,  the  two 
triangles  are  equal,  and  A  D  is  equal  to  B  C,  and  A  B 
equal  to  D  C. 

93.  Axiom,  *  If  one  end  of  a  straight  line  stands  still 
while  the  other  tunis  round,  the  end  that  moves  will  hegin 
to  move  in  a  direction  at  right  angles  to  that  of  the  line 
itself    Thus  if  A  B  were  to  ^ 

begin  to  turn  about  the  point    A B 

A,  B  would  hegin  to  move 

either  towards  C  or  towards  D.  [If  this  proposition  is  not 
acknowledged  as  an  axiom,  the  proof  is  in  Ex.  XIX.  at  the 
close  of  the  chapter.] 

94.  Theorem,  The  angles  of  a  triangle  cannot  be  al- 
tered without  altering  the  length  of  the  sides.  Proof.  If 
in  any  triangle,  as  ABC,  the  sides  were  unchangeable, 
any  alteration  of  the  angles  A  and  B  would,  by  article  93, 
make  the  point  C  move  in  two  directions  at  once,  (namely, 
at  right  angles  to  A  C,  and  at  right  angles  to  B  C,)  which 
is  impossible,  and  therefore  the  angles  cannot  be  altered. 


THE   PYTHAGOBEAN   PROPOSITION.  35 

95.  Corollary,  If  the  tlireo  sides  of  a  tri.in<j;l(^  m-o 
respectively  equal  to  tlic  tlnvo  sides  of  anotlier  triuiigle, 
the  angles  of  one  must  be  equal  to  those  of  the  other,  and 
the  equal  angles  are  enclosed  in  the  equal  sides. 

9G.  Theorem,  If  the  opposite  sides  of  a  quadrangle  are 
equal,  the  quadrangle  is  a  parallelogram.  Proof,  If,  in  the 
quadrangle  A  B  C  D,  the  sides  A  B  and  C  D  arc  equal, 
and  also  the  sides  A  D  and  B  C  are  equal,  then,  by  draw- 
ing tjic  diagonal  A  C,  we  have  the  triangles  ABC  and 
ADC  composed  of  equal  sides,  and,  by  article  95,  the 
angle  D  A  C  must  be  equal  to  the  angle  A  C  B,  and  tlie 
angle  D  C  A  to  the  angle  B  A  C ;  whence,  by  article  89, 
the  figure  is  a  parallelogram. 

97.  Theoreni,  The  area  of  a  rectangle  is  the  product 
of  its  length  by  its  breadth.  Proof,  By  drawing  lines,  at 
a  distance  apart  equal  to  the  unit 

of  length,  parallel  to  the  sides  of  the 

rectangle,  we  shall  (articles  87-89) 

divide    the     rectangle     into    little 

squares,  each  of  which  is  a  unit  of 

surface.     Moreover,  these   squares 

are  arranged  in  as  many  rows  as  there  are  units  of  length 

in  one  side  of  the  rectangle,  each  row  containing  as  many 

squares  as  there  are  units  of  length  in  the  other  side  ;  so 

tliat  the  whole  number  of  squares  is  found  by  multiplying 

the  length  of  the  rectangle  by  its  breadth. 

98.  Scholium,  In  the  above  proof  it  is  taken  for  grant- 
ed that  the  sides  of  the  rectangle  can  be  divided  into  units 
of  length.  This  can  usually  be  done  by  taking  the  units 
suihciently  short,  that  is  to  say,  if  the  lines  are  not  an  even 
number  of  inches  in  length,  we  may  take  tenths  of  an  inch 
as  the  unit ;  if  they  are  not  even  tenths,  we  can  divide  them 
into  hundredths,  or  thousandths,  or  even  millionths,  of  an 
inch.  If,  after  dividing  each  line  into  millionths  of  an  inch, 
any  thing  less  than  the  millionth  of  an  inch  were  left  at 


36  THE   PYTHAGOREAN   PROPOSITION. 

either  end,  it  would  be  too  small  to  be  taken  into  consid- 
eration. There  would  be  no  error,  even  in  reasoning,  from 
neglecting  it.  For  as  long  as  any  thing  is  left  at  the  ends 
of  the  lines,  we  can  choose  smaller  units ;  but  as  long  as  the 
units  are  of  any  size  at  all,  our  reasoning  holds  good,  and 
the  rectangle  is  measured  by  the  product  of  its  dimensions. 

99.  Theorem.  If  the  angles  of  one  triangle  are  equal 
to  those  of  another  triangle,  any  two  sides  of  one  of  the 
triangles  have  the  same  ratio 

to  each  other  as  that  of  the 
two  sides  including  the  same 
angle  in  the  other  triangle. 
Proof.  Let  the  triangles 
ABC  and  D  E  F  be  equi- 
angular with  respect  to  each 
other.  Place  the  vertex  A  upon  the  vertex  D,  and  allow 
the  side  A  B  to  fall  upon  the  side  D  E.  Since  the  angles 
A  and  D  are  equal,  the  line  A  C  will  fall  upon  the  line 
D  F ;  and  since  the  angles  C  and  F  are  equal,  the  line  B  C 
will  lie  parallel  to  the  line  E  F. 

Let  the  sides  A  B  and  D  E  be  divided  into  units  of 
length,  A  a,  a  B,  B  ^,  &c.  Through  the  points  of  division 
draw  lines  a  c,  h  c?,  &c.,  parallel  to  E  F.  Draw  also  the 
lines  a  6,  B/,  &c.,  parallel  to  D  F.  By  article  91,  the  tri- 
angles A  a  c,  «  B  e,  &c.,  are  equal.  By  article  92,  a  e  is 
equal  to  c  C,  B/*to  C  d^  &c.  Hence  it  is  easy  to  see  that 
A  B  is  composed  of  the  same  number  of  times  A  a,  that 
AC  is  of  A  c,  and  that  in  like  manner  D  E  is  as  many 
times  D  a,  as  D  F  is  times  D  c.  And  thus,  by  article  78, 
DE:DFi=:AB:DC,  because  each  of  these  ratios  is 
equal  to  B  a  :  6f  c. 

100.  Scholium.  If  the  lines  A  B  and  D  E  do  not  consist 
of  a  certain  number  of  times  the  first  unit  of  length  which 
we  have  chosen,  we  may  choose  a  unit  so  small  as  to 
make  the  remainder  small  enough  to  be  neglected. 


THE   PYTHAGOREAN   PROPOSITION. 


37 


101.  Definitions.  The  right  angle,  right  triangle,  legs, 
and  hypothenuse  are  defined  in  articles  14  and  17. 

10*2.  Theorem,  The  sum  of  the  three  angles  of  a  trian- 
gle is  equivalent  to  two  right  angles. 

This  proposition  has  been  proved  in  articles  26-31,  34- 
36,  and  57-62. 

103.  Corollary,  The  sum  of  the  two  angles  opposite 
to  tlie  legs  of  a  right  triangle  is  equivalent  to  one  right 
angle. 

104.  Corollary,  If  an  angle  opposite  a  leg  in  one  right- 
triangle  is  equal  to  an  angle  in  another  right  triangle,  the 
two  right  triangles  are  equiangular  with  respect  to  each 
other. 

105.  Theorem.  If  from  the  vertex  of  the  right  angle  in 
a  right  triangle,  a  line  be  drawn  at  right  angles  to  the 
hypothenuse  dividing  the  hypothenuse  into  two  segments, 
each  leg  is  a  mean  proportional  between  the  whole  hy- 
pothenuse and  the  segment  nearest  the  leg.  Proof,  Let 
ABC  be  a  right  triangle  with  aright  angle-at  C.  Draw 
C  F  at  right  angles  to  A  B.  The  triangle  B  E  C  is  right 
angled  at  E,  and  has  an  angle  at  B  equal  that  at  B  in  the 
triangle  ABC.  Hence,  by  article  104,  the  triangle  B  E  C 
has  its  angles  equal  to  those  of 
ABC.  Hence,  by  article  99, 
BE:BC  =  BC:BA.  In  the 
same  way  A  E  :  A  C  ;  :  A  C  : 
AB. 

106.  Theorem,  The  square 
on  the  hypothenuse  is  equiva- 
lent to  the  sum  of  the  squares 
on  the  legs.  Proof,  Let  A  C  B 
be  a  right  triangle,  with  a  right 
angle  at  C,  and  let  a  square 
be    drawn   on    each  side.      Draw 

4 


Fig.  A. 


C  F   at    right    angles 


38  THE    PYTIIAGOUEAN    riiOPOSlTIOX. 

to  A  B.  The  figure  B  F  will  be  a  rectangle,  because  all 
its  angles  will  be  right  angles.  It  w411,  therefore,  be  meas- 
ured by  the  product  of  B  E  into  E  F,  or  (since  E  F  :^:= 
B  G,  and  B  G  z=  B  A)  by  the  product  of  B  E  X  B  A. 
But  since  B  C  is  a  mean  proportional  between  the  lines 
B  E  and  B  A,  this  product  is  equal  to  B  C  X  B  C,  which 
is  the  measure  of  the  square  on  B  C.  That  is,  the  measure 
of  the  rectangle  B  F  is  the  same  as  that  of  the  square  on 
B  C.  In  the  same  manner  it  may  be  shown  that  the  rec- 
tangle A  F  is  equivalent  to  the  square  on  A  C.  But  the 
sum  of  these  two  rectangles  is  evidently  equal  to  the  square 
on  the  hypothenuse. 

107.  In  these  thirty-one  articles  I  have  given  you  a 
proof  of  the  Pythagorean  proposition  in  the  usual  synthet- 
ic form.  Parts  of  the  proof  are  not  completely  filled  out ; 
but  the  omitted  steps  are  so  short  and  easy,  that  I  think 
you  will  have  no  difficulty  wdiatever  in  supplying  them. 
Do  not  be  satisfied  with  understanding  each  of  the  thirty- 
one  articles,  but  examine  them  closely  from  the  76tli  to 
the  106th,  and  see  whether  I  have  introduced  any  thing 
which  is  not  necessary  to  the  proof  of  106.  In  making 
this  examination,  it  w^ill  be  most  convenient  for  you  to 
proceed  backward. 

These  thirty-one  articles  have  been  here  introduced  as 
lemmas,  i.  e.,  preparatory  propositions,  for  demonstrating 
the  Pythagorean  proposition.  But  they  are  also,  each  one, 
truths  worth  knowing,  and  will  aid  in  establishing  many 
theorems  that  have  no  connection  with  a  right  triangle. 

108.  Another  mode  of  analyzing  this  proposition  w^ould 
be  suggested  by  our  knowledge  of  the  fact  that  nny  trian- 
gle is  equivalent  to  half  a  rectangle  of  the  same  base  and 
altitude.  I  will  not  lead  you  through  this  analysis,  but 
w^ill  simply  build  up  for  you,  by  synthesis,  a 


THE    PYTHAGOREAN    TROPOSITION.  39 

Second  Proof  of  t/ie  Pythagorean  Proposition. 

109.  Definitions,  Any  side  of  a  triangle  or  quadrangle 
may  be  called  its  base,  and  the  altitude  of  the  figure  is  the 
distance  from  the  base  to  the  most  distant  vertex  of  the 
figure.  This  distance  is  measured  by  a  straight  line  at 
right  angles  to  the  base,  and  contained  between  the  vertex 
and  the  base,  prolonged  if  need  be. 

110.  Theorem,  Every  parallelogram  is  equivalent  to  a 
rectangle  of  the  same  base  and  altitude.  Proof,  Let 
A  B  C  D  be  a  rectan-  Fig.  c. 

gle,  and  A  B  E  F  a  par- 
allelogram having  the 
same  base  A  B,  and  the 
same  altitude  B  D.  It 
is  manifest  if  the  trian- 
gle B  D  F  by  which  the 
parallelogram  overlaps  the  rectangle  is  equal  to  the  trian- 
gle A  E  C  by  which  the  rectangle  overlaps  the  parallelo- 
gram, the  two  quadrangles  are  equivalent.  But  A  E  and 
its  adjacent  angles  are  equal  to  B  F  and  its  adjacent  angles, 
and  therefore  the  triangles  are  equal  (Art.  91),  and  the 
quadrangles  equivalent. 

111.  Theorem,  Every  triangle  is  equivalent  to  half  a 
rectangle  of  the  same  base  and  altitude.  Proof,  Let 
A  F  B  be  a  triangle,  and  A  B  C  D  a  rectangle  having  the 
same  base  A  B  and  the  same  altitude  B  D,  (Fig.  C.)  Con- 
tinue C  D  to  F,  and  draw  A  E  parallel  to  B  F.  The  tri- 
angle A  E  F  has  its  three  sides  equal  to  those  of  A  B  F  ; 
the  triangles  are,  therefore,  equal  to  each  other  (Art.  95) ; 
and  each  is  equal  to  half  the  parallelogram  A  B  E  F,  which 
is  equivalent  to  the  rectangle  A  B  C  D. 

112.  Theorem,  The  square  on  the  hypothenuse  is 
equivalent  to  the  suni  of  the -squares  on  the  legs.     Proof, 


40  THE    PYTHAGOREAN   PROPOSITION. 

Having  drawn  the  figure  (Fig.  A.),  as  for  the  former 
proof,  draw  the  lines  O  B,  and  B'  C.  The  triangle  A  B'  C 
has  the  same  base  A  B',  and  the  same  altitude  A  E  as  the 
rectangle  A  F,  and  is  equivalent  to  half  that  rectangle. 
The  triangle  ABC  has  the  same  base  A  C^  and  the  same 
altitude  A  C  as  the  square  O  C,  and  is  equivalent  to  half 
that  square ;  so  that  if  the  triangles  ABC  and  A  B'  C 
are  equal,  the  rectangle  is  equivalent  to  the  square.  But 
these  triangles  are  equal,  for  if  A  B'  C  were  turned  about 
the  vertex  A  as  on  a  pivot  until  the  point  C  covered  C, 
then  B'  would  cover  B,  and  the  triangles  would  coincide. 
For  A  C  would  rotate  through  a  right  angle,  and  A  B' 
through  a  right  angle ;  and  A  C  =  A  C,  and  A  B  =  A IV. 
113.  This  proof  of  the  Pythagorean  proposition  is  more 
strictly  geometrical  than  the  preceding,  as  it  does  not  in- 
volve the  idea  of  multiplying  lines  to  measure  areas.  But 
you  must  remember  that  both  are  equally  conclusive.  I 
have  here  also  omitted  some  of  the  shorter  steps.  You 
should  not  only  be  able  to  fill  out  these  steps  when  the 
omission  is  pointed  out  to  you,  but  also  to  discover  the 
omission  for  yourselves.  Take  the  proofs  which  I  have 
written  down  and  examine  them  step  by  step,  asking  at 
each  step.  Is  that  strictly  self-evident  ?  Can  it  be  ques- 
tioned ?  Can  it  be  divided  into  two  steps  ?  Is  there  need 
of  proof?  If  so,  has  the  proof  been  given  in  a  previous 
article?  It  is  only  by  such  an  earnest  study  of  the  book 
and  of  the  subject  that  you  can  make  the  process  of  math- 
ematical reasoning  become  a  sure  and  pleasant  road  for 
you  to  the  discovery  of  truth. 

Examples. 

Demonstrate  the  theorems  that  follow. 

VIII.  If  one  triangle  have  two  sides  and  the  included 
angle,  equal  to  two  sides  and  the  included  angle  in  another 
triangle,  the  two  triangles  are  equal. 


THE   PYTHAGOREAN   PROPOSITION.  41 

IX.  Lines  drawn  from  a  point  in  a  pei-pendiciilar  to 
points  at  equal  distances  from  its  foot  are  equal. 

X.  The  perpendicular  is  the  shortest  line  from  a  point 
^to  a  straight  line.     [This  may  be  proved  from  the  Pythag- 
orean proposition.] 

XI.  An  isosceles  triangle  is  one  in  which  two  sides  are 
equal.  Prove  by  VIII.  (having  first  drawn  one  line  in  the 
triangle)  that  the  angles  opposite  the  equal  sides  are  equal. 

XII.  An  equilateral  triangle  is  equiangular. 

XIII.  A  line  bisecting  (dividing  equally)  the  angle 
between  two  equal  sides  in  a  triangle,  is  perpendicular  to 
the  third  side,  and  bisects  it. 

XIY.  Two  angles  in  a  triangle  being  equal,  the  opposite 
sides  are  equal.  [Use  Art.  91,  looking  at  the  triangle 
from  both  sides  of  its  plane,  or  conceiving  it  turned  over.] 
*'  XV.   An  equiangular  triangle  is  equilateral. 

XVI.  If  one  side  of  a  triangle  is  prolonged,  the  external 
angle  thus  formed  is  equal  to  the  sum  of  the  opposite  in- 
terior  angles,  and  so  greater  than  either  of  them. 
7\.  XVII.  If  one  side  of  a  triangle  is  longer  than  another, 
the  angle  opposite  the  first  is  greater  than  that  opposite 
the  second.  [Let  ABC  be  the  triangle,  and  A  C  be 
longer  than  A  B.  Put  the  point  D  on  A  C,  making  A  D 
=r  A  B.  Join  B  D.  ISTow  prove  XVII.  by  means  of  XL 
and  the  last  clause  of  XVI.] 

XVIII.  In  an  isosceles  triangle  either  of  the  equal 
angles  is  the  complement  of  half  the  third  angle. 

XIX.  If  a  straight  line,  A  B,  revolves  about  the  point 
A,  the  point  B  moves  at  right  angles  to  B  A.  [Allow  B 
to  have  moved,  complete  the  triangle,  apply  XVIIL,  and 
then  suppose  the  distance  moved  to  be  infinitesimal.] 

XX.  The  square  on  the  diagonal  of  a  square  is  double 
that  square. 

XXL  If  the  four  angles  of  a  quadrangle  consist  of  two 
pairs  of  opposite  equal  angles,  the  quadrangle  is  a  paral- 
lelogram. 4  * 


42  THE    MAXIMUM    AREA. 

XXII.  The  sum  of  tlie  nngles  of  a  quadrangle  is  equiv- 
alent to  four  right  angles ;  of  a  pentagon,  to  six ;  of  a 
heptagon,  to  ten. 

XXIII.  The  perpendicular  on  the  hypothenuse  froni 
the  opposite  vertex  is  a  mean  proportional  between  the 
segments  of  the  hypothenuse. 

XXIV.  The  areas  of  triangles  having  the  same  base 
are  proportional  to  their  altitudes  5  that  is,  have  the  ratio 
of  their  altitudes. 


CHAPTER    VIII. 


THE    MAXIMUM    AREA. 


114.  I  WILL  prove  only  one  more  proposition;  but  I  will 
select  a  difficult  one,  in  order  that  it  may  require  a  num- 
ber of  preliminary  proofs.  I  will  select  the  jiroposition 
given  in  the  "  First  Lessons  in  Geometry,"  Chap.  XXIII. 
§  14 :  Of  all  isoperimetrical  figures  the  circle  is  the  very 
largest. 

115.  When  we  attempt  to  analyze  this,  we  shall  see 
that  it  implies  that  any  regular  polygon  is  less  than  a  cir- 
cle isoperimetrical  with  it,  and  that  any  other  polygon  is 
less  than  a  regular  one  isoperimetrical  with  it. 

116.  Let  us  begin,  however,  by  defining  a  few  of  the 
Avords  we  shall  need  to  use. 

117.  A  polygon  is  a  plane  figure  bounded  by  straight 
lines. 

118.  The  perimeter  of  a  polygon  is  the  sum  of  the  length 
of  its  sides. 

119.  Isoperimetrical  polygons  are  those  of  equal  perim- 
eter. 

120.  Among  quantities  of  the  same  kind,  the  largest  is 
called  a  maximum. 


THE    MAXIMUM   AREA. 


43 


If  the  figure 


121.  A  circle  is  a  plane  figure,  bounded  by  one  line  that 
curves  equally  in  every  part.  This  line  is  called  the  circum- 
ference of  the  circle,  and  frequently  the  line  itself  is  called 
the  circle.     Portions  of  the  circumference  are  called  arcs.    ' 

12*2.  Theorem.  There  is  a  point  within  the  circle 
equally  distant  from  every  point  of  the  circumference. 
This  })oint  is  called  the  centre  of  the  circle. 
Vroof,  Let  A  D  and  B  D  be  equal  adja- 
cent arcs  in  a  circumference.  Through  the 
points  A,  B,  and  D  draw  lines  at  right  an- 
gles to  the  curve  at  those  points.  Now, 
since  the  circle  curves  uniformly  at  every 
point,  B  D  is  in  all  respects  equal  to  D  A. 
BCD  were  laid  upon  D  C  A  the  arcs  would  coincide ; 
and  also,  D  C  would  go  in  the  direction  of  A  C,  while  B  C 
vv  ould  go  in  the  present  direction  of  D  C.  The  figures 
A  C  D  and  D  C  B  would  thus  coincide,  and  A  C  =  D  C 
=rr  B  C.  Hence  the  points  A,  B,  and  D  are  equally  dis- 
tant from  C.  But  A  and  B  may  be  taken  any  where  in 
the  circle,  only  provided  they  are  e quail y#dist ant  from  D ; 
and  hence  every  point  in  the  circle  is  equally  distant  from 
C,  the  centre  of  the  circle. 

123.  A  straight  line  joining  the  centre  to  the  circum- 
ference is  called  a  radius.  The  straight  line  formed  of  two 
opposite  radii  is  called  a  diameter. 

124.  All  radii  are  of  course  equal  to  each  other,  and  all 
diameters  equal  to  each  other. 

125.  A  straight  line 
joining  the  two  ends 
of  an  arc  is  called  a 
chord. 

126.  A  straight  line 
which,  however  much 
prolonged,  touches  the 
circle  in  one  point  only,  is  called  a  tangent. 


44  THE    MAXIMUM   AREA. 

127.  It  is  manifest  that  the  tangent  coincides  in  direc- 
tion with  the  arc  at  the  point  of  contact. 

128.  A  polygon  formed  wholly  of  chords  in  a  circle  is 
said  to  be  inscribed  in  that  circle. 

129.  A  polygon  formed  wholly  by  tangents  to  a  circle 
is  said  to  be  circumscribed  about  the  circle. 

130.  The  circle  is  said  to  be  inscribed  in  the  circum- 
scribed polygon,  and  to  be  circumscribed  about  the  in- 
scribed polygon. 

131.  If  a  polygon,  about  which  a  circle  can  be  circum- 
scribed, or  in  which  a  circle  can  be  inscribed,  has  its  sides 
equal,  one  to  the  other,  the  polygon  is  called  a  regular 
polygon,  and  the  centre  of  these  circles  is  called  also  the 
centre  of  the  polygon. 

132.  Let  us  now  attempt  to  analyze  the  proposition 
that  the  circle  is  the  maximum  among  isoperimetrical  pol- 
ygons. This  is  equivalent  to  saying  that  if  an  isoperimet- 
rical circle  and  regular  polygon  are  laid  one  over  the  other, 
the  polygon  will  be  the  smaller.  But  we  see  that  by  lay- 
ing them  one  oii  the  other,  a  circle  inscribed  in  the  poly- 
gon would  be  smaller  than  the  isoperimetrical  circle.  The 
question  of  course  suggests  itself,  whether  the  area  of  a 
regular  polygon  is  not  proportional  to  the  radius  of  the 
inscribed  circle.  Now,  it  is  plain  that  it  is.  For  by  divid- 
ing the  polygon  into  triangles,  by  lines  from  its  centre  to 
its  vertices,  we  find  the  area  of  the  polygon  will  be  the 
sum  of  the  areas  of  the  triangles,  and  these  areas  will  be 
measured  by  half  the  product  of  the  perimeter  multiplied 
by  the  radius  of  the  inscribed  circle.  The  area  of  the 
isoperimetrical  circle  will  be  measured  by  half  the  product 
of  the  perimeter  multiplied  by  the  radius.  But  as  the 
perimeters  of  the  polygon  and  the  isoperimetrical  circle  are 
the  same,  and  the  radius  of  tlie  inscribed  circle  is  smaller 
than  that  of  the  isoperimetrical  circle,  it  is  evident  that  the  , 
area  of  the  polygon  is  smaller  than  that  of  the  isoperimet- 
rical circle. 


THE    MAXIMUM   AREA.  45 

It  will  now  remain  to  show  that  the  area  of  a  regular 
polygon  is  greater  than  that  of  an  isoperimetrical  irregular 
polygon.  It  is  evident  that  this  can  be  done,  since  a  pol- 
ygon of  given  sides  is  manifestly  largest  when  most  nearly 
circular,  and  a  polygon  of  a  given  number  of  sides  is  man- 
ifestly largest  when  the  sides  are  equal.  We  can  surely 
have  no  difficulty  in  proving  these  two  points,  and  then 
our  proof  will  be  complete. 

133.  Let  us  return,  then,  to  the  synthetic  mode,  and 
establish  these  propositions :  First,  that  the  maximum  of 
polygons  formed  of  given  sides  may  be  inscribed  in  a  cir- 
cle ;  secondly,  that  the  maximum  of  isoperimetrical  poly- 
gons having  a  given  number  of  sides  has  its  sides  equal ; 
and  thirdly,  that  such  a  regular  polygon  is  of  smaller  area 
than  a  circle  isoperimetrical  with  it. 

134.  Theorem,  The  area  of  a  triangle  is  found  by  mul- 
tiplying the  base  by  half  the  altitude.  This  theorem  has 
been  already  proved.     (Art.  111.) 

135.  We  shall  need  the  Pythagorean  proposition,  which 
implies  all  the  propositions  into  which  we  have  already 
analyzed  it.     (Arts.  64-113.) 

136.  Theorem.  Of  two  unequal  lines,  from  a  point  to  a 
third  straight  line,  the  shorter  is  more  near^y  perpendicu- 
lar to  the  third  line.  Proof,  Let  C  be  the  given  point, 
and  AD  the  third  straight 
line.  Let  C  A  and  C  B  be 
two  lines,  of  which  C  B  is 
the  shorter.  Draw  C  D  per- 
pendicular to  A  D.  We  wish 
to  prove  that  B  D  is  shorter  than  A  D.  But  this  is  mani- 
fest from  the  Pythagorean  proposition,  since  the  square  on 
A  D  is  the  difference  of  the  squares  on  ,K  C  and  C  D, 
and  the  square  on  B  D  is  the  difference  of  the  squares  on 
B  C   (which  is  smaller  than  A  C)  and  the  same  C  D. 

137.  Corollary,  A  perpendicular  is  the  shortest  line 
from  a  point  to  a  given  straight  line. 


46  THE    l^tAXIMUM   AREA. 

188.  Theorem,  The  radius  is  perpendicular  to  the  tan- 
gent at  its  extremity.  Proof,  For  if  not,  then,  by  Art.  137, 
the  tangent  would  pass  inside  the  circle,  which  is  contrary 
to  its  definition. 

139.  Corollary,  The  radius  is  perpendicular  to  the  arc 
at  its  extremity. 

140.  Theorem,  Either  side  of  a  triangle  is  shorter  than 
the  sitm  of  the  other  two.  Proof,  Upon  either  side,  pro- 
longed if  necessary,  drop  from  the  opposite  vertex  a  per- 
pendicular. The  sum  of  the  distances  from  the  foot  of  this " 
perpendicular  to  the  adjoining  vertices  cannot  be  less  than 
the  whole  of  the  selected  side,  but  must,  by  187,  be  less 
than  the  sum  of  the  other  two.  Another  proof.  The 
straight  line  is  the  shortest  line  between  its  extremities. 

141.  Theorem,  The  maximum  of  triangles  having  two 
sides  given  is  formed  when  these  two  sides  are  at  right 
angles.  Proof,  JLet  A'  B, 
A  B,  and  A'^  B  be  equal  to 
each  other.  The  area  of 
A'BC,  ABC,  orA^'BC, 
being  found  by  multiplying 
B  C  into  half  the  perpen- 
dicular height  of  A,  A',  or 
A'',  above  B  C,  will  be  in  proportion  to  that  height. 
Let,  then,  A  B  be  perpendicular  to  B  C,  and  the  height 
of  A  above  the  base  will  equal  B  A.  But  the  height  of 
A"  above  the  base  must,  by  187,  be  less  than  B  A'',  which 
is  equal  to  B  A. 

142.  An  angle  is  said  to  be  measured  by  an  arc  of  a 
circle  such  as  would  be  intercepted  by  radii  making  that 
angle  with  each  other.  And,  since  the  circumference 
curves  equally  in  all  parts,  and  the  radii  are  at  right  angles 
to  it,  it  is  evident  that  this  measure  is  just,  and  that  the 
angle  will  bear  the  same  ratio  to  four  right  angles  that 
the  arc  bears  to  a  whole  circumference,  whatever  be  the 
size  of  the  circle. 


THE   MAXIMUM   AREA.  47 

143.  Theorem,  If  two  sides  in  a  triangle  are  equal,  the 
angles  opposite  those  sides  -p 

are    equal.      Proof,     Let  ^^""'''^T^*^'"*"^ 

A  B  and  B  C  be  equal  sides      }l^^^^^^         •  ^^"^^^  C 

in    a    triangle.      Imagine 

A  C  divided  in  the  centre,  at  the  point  h.  The  triangles 
A  B  Z>  and  C  B  ^  will  now  be  composed  of  equal  sides,  and 
we  have  already  proved  (Arts.  91-95)  that  they  must  have 
equal  angles  ;  that  is,  the  angle  at  A  is  equal  to  that 
at  C. 

144.  Theorem,  If  one  side  of  a  triangle  is  prolonged, 
the  external  angle  is  equal  to  the  sum  of  the  opposite  in- 
ternal angles.     This  has  been  proved  in  Art.  57. 

145.  Two  chords  starting  from  one  point  in  a  circum- 
ference intercept  double  the  arc  that  would  be  intercepted 
by  radii  making  the  same  angle ;  that  is,  the  angle  of  the 
chords  is  measured  by  half  the  arc  included  between  them. 
Proof,  If  one  chord,  as  AB,  passes 
through  the  centre  D  of  the  circle,  it  is 
plain  that  by  drawing  D  C  the  angle 
C  D  B  will  be  equal  to  the  sum  of  the 
angles  CAD  and  D  C  A.  But  since 
D  A  and  D  C  are  equal,  these  angles 
arc  equal,  and  C  D  B  is  equal  to  twice  CAD. 

If  neither  chord  passes  through  the  centre  of  the  circle, 
we  can  draw  a  third  chord,  starting  from  A,  passing 
through  the  centre  of  the  circle,  and  apply  this  reasoning 
to  the  two  angles  formed  with  this  third  chord  by  the 
other  two.  The  angle  of  the  other  two  chords  will  simply 
be  the  sum  or  the  difference  of  these  two  angles. 

146.  Corollary,  If  the  vertex  of  a  right  angle  be  placed 
in  the  circumference,  the  sides  will  intercept  a  semicircle. 

147.  Corollary,  If  a  circle  be  circumscribed  about  a 
triangle,  and  one  side  of  the  triangle  passes  through  the 
centre  of  the  circle,  the  opposite  angle  is  a  right  angle. 


48 


THE   MAXIMUM  AREA. 


148.  Theorem.  The  maximum  of  polygons,  having  all 
the  sides  given  but  one,  may  have  a  circle  circumscribed 
about  it,  having  the  unknown  side  for  a  diameter. 
Proof.  Let  A  B  C  D  E  be  the  maxi- 
mum polygon,  formed  of  given  sides 
A  B,  B  C,  &c.,  and  the  unknown  side, 
A  E.  Join  B  E  by  a  straight  line. 
Now,  since  the  polygon  is  a  maximum, 
we  cannot,  leaving  B  E  unaltered, 
by  altering  A  E  enlarge  the  triangle 
ABE,  because  that  would  enlarge 
the  polygon.  The  angle  ABE  is  therefore  a  right  angle, 
by  Art.  141,  and  a  circumference,  having  A  E  for  its  di- 
ameter, would  pass  througli  the  point  B.  In  like  manner 
it  can  be  shown  that  a  circumference  having  the  same 
diameter  Svould  pass  through  each  of  the  other  points. 

149.  Theorem.  The  maximum  of  polygons  formed  with 
given  sides  can  be  inscribed  in  a  circle.  Proof,  Let 
A  B  C  D  E  be  a  polygon,  formed  of 
given  sides,  with  a  circle  circumscribed 
about  it.  Draw  the  diameter  A  F,  and  ' 
join  F  C  and  F  D.  The  polygons 
A  B  €  F  and  A  E  D  F  are  now  maxi- 
mum polygons,  and  therefore  ABODE 
must  also  be  a  maximum,  since  its  en- 
largement would  enlarge  the  sum  of 
the  other  two. 

We  have  thus  proved  the  converse  of  the  proposition, 
and  the  proposition  is  true,  unless  there  is  more  than  one 
maximum  form  of  the  polygon. 

The  converse  is  more  easily  proved  than  the  proposition, 
and  I  therefore  proved  it,  on  the  assumption  that  there  is 
but  one  maximum  form.  That  is,  I  have  proved  that  a 
polygon  of  given  sides,  when  inscribed  in  a  circle,  is  a 
maximum ;  but  that  does  not  strictly  prove  that  the  maxi- 


THE    MAXIMUM   AREA.  49 

mum  can  always  be  inscribed  in  a  circle ;  except  on  the 
assumption,  which  is,  however,  a  safe  one,  that  a  polygon 
formed  of  given  sides,  arranged  in  a  given  order  of  succes- 
sion, can  have  but  one  maximum  form. 

150.  Theorem,  Of  isopcrimetrical  triangles  with  one 
side  given,  the  maximum  has  the  two  undetermined  sides 
equal.  Proof,  In  order  to  prove 
this  we  have  only  to  show  that 
the  point  A  is  at  its  greatest  dis- 
tance from  the  base  B  C,  w^hen 
opposite  the  middle  of  it.  This 
might    seem    scarcely  to    need 

proof  For  when  we  use  a  string  and  stick  to  illustrate 
the  problem,  we  can  see  that  by  sliding  the  finger  from 
the  middle  of  the  string,  it  can  be 
brought  down  into  a  line  with  the 
stick;  and  the  greatest  height  from  the 
stick  is  near  the  middle  of  the  string. 
Further  consideration  shows  it  must 
be  exactly  at  the  centre  of  the  string, 
because  the  finger  and  string  have  precisely  the  same  rela- 
tion to  one  end  of  the  stick  as  to  the  other ;  and-  a  motion 
towards  either  end  must  afiect  the  height  of  the  finger  in 
a  similar  manner. 

This  reasoning  is  doubtless  satisfactory  to  every  fair 
mind.  Yet  it  is  not  a  good  mathematical  demonstration, 
and  I  have  given  it  to  you  for  the  purpose  of  illustrating 
the  peculiar  nature  of  mathematical  reasoning.  The  rea- 
soning just  given  leaves  no  real  doubt  on  the  mind,  but  it 
is  rather  because  we  see  with  the  eye  that  the  finger  is 
highest  in  the  middle,  than  because  we  see  with  the  mind 
that  it  must  be.  There  is  another  step  still  lacking,  to 
prove  to  us  that  the  highest  points  are  not  on  each  side 
of  the  exact  middle,  as  that  would  satisfy  the  conditions 
of  symmetry  and  of  declination  towards  each  end.    Let  us 


50  THE  MAXIMUM   AREA. 

then  seek  a  proof  which  shall  not  force  us  to  consider  the 
whole  motion  of  the  finger,  but  which  shall  simply  compare 
two  forms  of  the  triangle,  one  with  the  finger  in  the  mid- 
dle of  the  string,  and  one  with  the  finger  on  one  side,  ^y 

151.  Theorem,  If  a  straight  line  be  drawn  from  the 
vertex  of  two  equal  sides  in  a  triangle,  at  right  angles  to 
the  third  side,  it  divides  the  third  side  into  equal  parts. 
Proof,    Let  c  and  a  be  ^ 

equal   sides   in   a  triangle  e^.^''^'T^'*^**-^ 

ABC.  Since  the  angles  .^^^^^"^  {  ^^""""^^^^ 
at  A  and  C  are  equal,  the  ^ 

angles  SBC  and  ^  B  A  are  also  equal.  If,  therefore, 
the  triangle  B  5  C  be  folded  over  on  the  line  B  ^,  the  line 
a  will  take  the  same  direction  as  the  line  c,  and,  being  of 
the  same  length,  will  coincide  with  it.  Hence,  b  C  will 
also  coincide  with  h  A,  and  the  two  lines  must  be  of  equal 
length. 

•152.  Mio  2'>roof  of  Art,  150.  Let  A  B  C  and  A  B  C  be 
isoperimetrical,  and  let  A  B  and  B  C  be  equal.  Continue 
A  B  to  D,  making  B  D  =:  B  A  rzr:  B  C, 
and  join  D  C.  Then,  by  Art.  147, 
the  angle  D  C  A  is  a  right  angle. 
Draw  B'  E  making  it  equal  to  B'  C. 
Join  A  E.  A  E  will  be  less  than 
the  sum  of  A  B'  and  B'  E,  that  is,  less 
than  A  B'  and  B'C,  that  is,  less  than    ^  ^ 

A  B  and  B  C,  that  is,  less  than  A  D.  But  if  A  E  is  less 
than  A  D,  then  C  E  must  be  less  than  C  D,  by  Art.  136. 
Draw  B  H  and  B'  I  at  right  angles  to  CD;  we  have  C  I, 
which  is  half  C E,  less  than  C  H  which  is  half  CD.  But 
C  I  and  C  H  are  the  altitudes  of  the  triangles  ABC  and 
A  B'  C  above  their  common  base  A  C.  The  triangle  with 
the  undetermined  sides  equal  has  the  greatest  altitude, 
and  must  be  the  largest  triangle. 

153.  Theorem,    The  maximum  of  isoperimetrical  poly- 


A 

I> 

/) 

E 

^<C;^i. 

H 

/^■f\ 

I 

THE  MAXIMUM  AREA.  51 

gons  of  a  given  number  of  sides  is  equilateral,  that  is,  lias 
equal  sides.  Proof.  Let  A  B  C  E  D 
be  the  maximum  of  isoperimetrical 
polygons  of  a  given  number  of  sides. 
Then  A  B  must  equal  B  C.  For  if  it 
did  not,  then  after  joining  A  and  C 
we  could  enlarge  the  triangle  ABC 
by  equalizing  A  B  and  A  C,  and  thus 
enlarge  the  polygon  without  altering 
the  number  of  sides  of  the  perimeter, 
and  the  present  form  would  not  be  the  maximum. 
In  like  manner  we  may  prove  that  B  C  =  C  E,  &c. 

154.  Corollary,  The  maximum  of  isoperimetrical  poly- 
gons of  a  given  number  of  sides  is  regular  by  Arts.  149 
and  153. 

155.  Axiom,  A  circle  may  be  considered  as  a  regular 
polygon  having  an  unlimited  number  of  sides.  And  this 
regular  polygon  may  be  considered  as  either  inscribed  in 
or  circumscribed  about  the  real  curve. 

156.  Theorem.  The  area  of  a  regular  polygon  is  meas- 
ured by  half  the  product  of  the  perimeter  into  the  radius 
of  the  inscribed  circle.  Proof.  Eor  if  lines  be  drawn  from 
each  vertex  of  the  polygon  to  the  centre,  the  polygon  will 
be  divided  into  triangles  having  a  common  altitude  equal 
to  the  radius  of  the  inscribed  circle,  the  sum  of  the  bases 
of  these  triangles  being  equal  to  the  perimeter  of  the 
polygon.       yf^. 

157.  Corollary.  The  area  of  a  circle  is  measured  by 
half  the  product  of  the  radius  into  the  circumference. 

158.  Theorem.     The  perimeter  of  a  circum-  A 
scribed  polygon   is   greater  than   the   circum- 
ference of  the  circle.     Proof  Let  A  B  be  half  ^\ 
a  side   of  a  circumscribed  polygon,  and  D  B 
the  portion  of  arc  intercepted  by  lines  drawn 
from  A  and  B  to  the  centre  of  the  circle.    Di- 


52  THE   MAXIMUM   AEEA. 

vide  D  B  into  arcs  so  small  that  each  may  be  considered 
as  a  short  straight  line.  Through  the  points  of  division 
draw  lines  extending  from  the  line  A  B  to  the  point  C. 
At  the  end  B,  the  little  arcs  are  equal  to  the  correspond- 
ing pieces  of  the  line  A B;  but  as  you  approach  A  the 
divisions  of  the  line  grow  longer  than  the  correspond- 
ing divisions  of  the  arc,  for  two  reasons ;  first,  the  little 
arcs  are  at  right  angles  to  the  radii,  while  the  portions 
of  the  line  are  not  (Art.  139) ;  secondly,  the  little  arcs  are 
nearer  to  the  point  C,  towards  which  the  radii  converge. 
The  whole  of  A  B  must  therefore  be  longer  than  the  whole 
of  D  B.  But  it  is  manifest  that  the  circumference  consists 
of  as  many  times  D  B  as  the  perimeter  does  of  the  line 
AB. 

159.  Corollary,  The  circle  inscribed  in  a  regular  poly- 
gon is  smaller  than  a  circle  isoperimetrical  with  the  poly- 
gon, and  has  a  shorter  radius. 

160.  Corollary,  The  circle  is  the  maximum  among 
isoperimetrical  regular  polygons. 

161.  Corollary,  The  circle  is  the  maximum  among  iso- 
perimetrical figures ;  a  proposition  towards  which  we  have 
been  directing  our  course  through  48  articles,  some  of 
which  are  themselves  complex  propositions,  referring  to 
the  preceding  chapters.  No  other  science  requires  any 
thing  like  such  long  trains  of  connected  reasoning  as  those 
used  in  the  mathematics.  An  argument  in  other  matters 
usually  consists  of  only  a  few  steps  —  what  are  called 
long  arguments  being  really  a  collection  of  shorter  inde- 
pendent proofs  of  the  same  thing.  In  the  mathematics, 
we  are  frequently  required  to  take,  as  in  the  present  in- 
stance, hundreds  of  consecutive  steps  to  attain  a  single 
position. 

162.  Scholium,  A  slight  modification  of  the  reasoning 
in  Arts.  158-160,  would  show,  that  of  isoperimetrical  pol- 
ygons that  is  greatest  which  has  the  greatest  number  of 
sides. 


THE    MAXIMUM    AREA.  53 

163.  He  that  really  wishes  to  learn  geometry  must  learn 
to  work  alone.  I  advise  the  learner  now  to  take  up  "  First 
Lessons  in  Geometry,"  and,  beginning  with  the  fourth 
chapter,  go  through  to  the  twenty-sixth,  trying  how  many 
of  the  facts  he  can  prove.  I  think  he  can,  if  he  sets  him- 
self to  work  with  a  good  will,  prove  the  greater  part. 
Perhaps  he  will  be  obliged  to  ask  some  help  of  his  teacher, 
but  I  think  not  much.  He  will,  however,  do  well  to  show 
liis  demonstrations  to  his  teacher  for  his  criticism. 

When  he  comes  to  Chap.  XXVI.  of  the  "  First  Lessons  ^ 
he  will  be  obliged  to  lay  down  the  book  again,  as  the 
propositions  in  the  remainder  of  the  book  cannot  be  proved 
without  the  aid  of  higher  branches  of  mathematics  —  Al- 
gebra, Trigonometry,  and  the  Calculus. 

164.  In  proving  the  Pythagorean  proposition,  and  the 
proposition  that  the  circle  is  the  maximum  among  isoperi- 
metrical  plane  figures,  I  have  tried  to  give  good  examples 
of  the  mathematical  mode  of  proof,  —  the  analysis,  in  which 
the  mind  turns  the  proposition  over  in  every  form,  trying 
all  sorts  of  experiments  upon  it  intellectually,  to  discover 
its  vulnerable  side,  —  and  the  synthesis,  by  which  we  then 
enter  step  by  step  into  the  very  secret  of  the  mystery. 

Analysis  consists  in  taking  the  proposition  itself  as  the 
starting  point,  and  going,  step  by  step,  to  self-evident 
truths,  or  at  least  to  truths  already  proved.  Synthesis 
consists  in  starting  with  self-evident  truths,  or  truths  al- 
ready proved,  and  going  step  by  step  to  the  truth  which 
you  would  prove.  But  synthesis  generally  requires  a  pre- 
vious rough  analysis,  by  which  you  select  the  proper  point 
of  departure  for  your  synthetical  reasoning. 

A  species  of  analysis,  called  rediictio  ad  ahsurdum^  is 
often  used  in  cases  where  true  analysis  or  true  synthesis 
is  difficult.  In  this  form  of  proof  you  assume  that  your 
proposition  is  not  true,  and  by  analysis  show  that  this 
would  lead  you,  step  by  step,  to  the  denial  of  self-evident 


54  •  THE    MAXIMUM   AREA. 

truth.  This  shows  the  proposition  to  be  true,  by  simply 
sljowing  that  it  cannot  be  false.  Art.  138  gives  an  in- 
stance of  this  proof;  also  Example  VI. 

I  think  you  will,  after  mastering  this  book  thoroughly, 
be  able  to  read  any  of  the  books  on  geometry  which  you 
will  be  at  all  likely  to  meet  with. 


GEOMETRICAL   CONSTRUCTION.  55 


PART  II. 
CHAPTER    I. 

GEOMETRICAL   CONSTRUCTION. 

1G5.  The  first  reason  for  learning  Geometry  is,  that  it 
toadies  us  truth.  This  reason  would  be  sufficient  in  itself. 
It  is  as  important  for  us  to  learn  truth  as  it  is  for  us  to  eat 
food.  But  there  is  another  reason  for  learning  Geometry, 
and  that  is,  the  use  which  we  may  make  of  its  truths. 

1G6.  The  uses  which  can  be  made  of  Geometry  are  of 
two  kinds.  We  can  use  it  in  the  investigation  of  other 
kinds  of  truths,  that  is,  in  studying  Astronomy,  Mechanics, 
Chemistry,  and  other  sciences ;  or  we  can  use  it  in  arts  and 
trades.  I  shall  not,  however,  attempt  to  keep  up  this  dis- 
tinction in  the  following  pages. 

1G7.  Definition,  We  can  frequently  solve  a  mathe- 
inatical  question  by  representing  given  quantities  as  lines 
and  angles,  constructing  or  drawing  the  figure  on  paper, 
and  then  measuring  the  lines  or  angles  representing  the 
unknown  quantities.  This  is  called  "  solution  by  geomet- 
rical construction."  I  will  explain  it  more  fully  a  few 
pages  farther  on. 

168.  The  first  requisite  for  geometrical  constructions, 
after  a  supply  of  drawing  paper  and  pencils,  and  a  plane 
table,  is  a  straight  ruler.  It  is  not  necessary  that  the  table 
should  be  perfectly  plane,  but  it  must  be  smooth,  and 
nearly  plane.  But  the  ruler  should  have  a  perfectly 
straight  edge;  at  least  as  nearly  so  as  the  material  of 
which  it  is  made  will  allow. 


56  GEOMETRICAL   CONSTRUCTION. 

The  mechanical  means  of  obtaining  a  straight  edge  will 
illustrate  the  uses  of  geometrical  knowledge.  The  axiom 
that  the  shortest  distance  between  two  points  is  measured 
by  a  straight  line,  shows  that  a  stretched  thread #will 
mark  a  straight  line,  and  afford  a  guide  for  sawing  out  a 
tolerably  straight  ruler.  The  sawn  edge  will  not,  howev- 
er, be  smooth  ;  and  in  the  process  of  smoothing  it  may  be 
brought  more  nearly  into  a  perfectly  straight  line  by  the 
application  of  various  tests. 

The  first  method  is  by  "sighting"  the  edge;  that  is, 
looking  at  it  with  one  end  very  near  one  eye,  and  observ- 
ing whether  the  farther  end  will,  upon  raising  the  near  end 
a  little  higher,  disappear  at  the  same  moment  that  the 
whole  edge  disappears.  If  any  part  of  the  edge  remains 
in  sight  after  any  other  part  has  disappeared,  the  edge 
cannot  be  perfectly  straight.  This  assertion  is  founded  on 
the  assumption  that  light  moves  in  straight  lines.  "  Sight- 
ing "  does  •  not  afford  a  very  delicate  means  of  testing  a 
straight  edge,  partly  on  account  of  the  impossibility  of 
looking  at  a  near  and  at  a  distant  point  at  the  same 
instant. 

A  second  test  is  founded  on  the  obvious  truth,  that  two 
lines  cannot  coincide  unless  they  are  both  straight  or  both 
have  the  same  bondings.  If  two  straight  edges  are  placed 
together,  and  touch  throughout  their  wliole  extent,  the 
probability  is  very  strong  that  they  are  perfectly  straight. 
That  probability  is  still  further  increased  if  they  continue 
to  touch  in  their  whole  extent  when  one  is  made  to  slide 
backwards  and  forwards  upon  the  other.  In  that  case 
they  must  be  either  straight,  or  else  arcs  of  the  same  cir- 
cumference. This  may  be  finally  tested  by  drawing  a  fine 
line  by  means  of  one  of  the  suspected  straight  edges  upon 
firm  paper,  and  then  applying  the  same  straight  edge  to 
the  opposite  side  of  the  line.  If  curved,  this  reversion 
will  at  once  show  it. 

169.  The  second  most  important  requisite  for  geomet- 


GEOMETRICAL   CONSTRUCTION. 


57 


rical  construction  is  a  pair  of  compasses,  or  dividers.  These 
are  made  of  various  degrees  of  delicacy,  and 
are  of  various  prices.  Some  have  merely  steel 
points,  by  which  circles  can  be  scratched 
upon  paper  or  upon  wood ;  othei*s  are  ar- 
ranged to  carry  lead  pencils,  or  to  cariy  ink 
in  a  peculiar  kind  of  pen.  The  best  ink  for 
such  uses  is,  however,  made  by  rubbing  the 
solid  "  Indian  ink"  with  water. 

170.  Compasses,  or  dividers,  have  two  uses,  as  their  two 
names  imply.  They  can  be  opened  to  any  width,  and  thus 
be  made  to  measure  the  length  of  lines,  and  the  distances 
between  points.  Or,  having  been  opened  to  the  width  of 
the  radius  of  a  required  circle,  one  point  can  be  held  still 
at  the  centre  while  the  other  traces  the  circumference. 
The  joint  should  be  finn  enough  to  prevent  the  radius 
from  readily  changing  its  length. 

171.  The  third  requisite  is  a  scale.  This  is  a  piece  of 
wood,  bone,  ivory,  or  metal,  marked  on  its  various  sides 
with  lines  at  equal  distances  (of  an  inch,  half  inch,  or  other 
convenient  unity),  having  at  one  end  also  a  diagonal  net- 
work for  measuring  tenths,  hundredths,  and  thousandths 
of  the  unit. 

The  lines  A  B  and  C  D  are  parallel,  and  one  unit  apart. 
This  space  is  divided  into  tenths  on  A  C  and  also  on  B  D, 
and  each  point  of 
division  on  A  C  is  ^ 
joined  to  the  next 
higher  point  on 
BD.  The  space  I 
between  the  line 
A  C  and  the  line 
D  B  is  divided 
into  •  tenths  by 
parallel  lines,  such 
as  4,  4,  and  5,  5.     The  modes  of  using  this   scale   will 


58  GEOMETRICAL   CONSTRUCTION. 

be  obvious  on  the  slightest  reflection.  The  distance  be- 
tween A  and  6  is  six  tenths  of  the  units,  between  B  and 
7  is  seven  tenths.  But  the  lines  A  B,  and  6,  7,  are,  at  any 
intermediate  point,  at  an  intermediate  distance  apart;  as, 
for  instance,  their  distance  measured  on  the  line  4,  4,  would 
be  64  hundredths  of  a  unit ;  measured  on  the  line  5,  5,  it 
would  be  65  hundredths ;  and  measured  at  three  tenths 
of  the  way  down  from  4,  4,  to  5,  5,  it  would  be  643  thou- 
sandths of  a  unit.  Thus,  with  very  fine  pointed  dividers 
you  can  readily  measure  to  the  thousandths  of  a  unit ;  or 
rather  measure  to  the  hundredths,  and  estimate  very  accu- 
rately to  thousandths.  If  you  have  no  such  scale  prepared, 
you  can,  with  a  very  fine  pencil  point,  sharpened  flat,  draw 
one  for  yourself  on  card,  and  make  it  durable  by  sizing  it 
with  a  drop  of  gum  water. 

172.  I  can  now  give  you  an  illustration  of  the  definition 
in  Art.  167.  Suppose  that  you  wish  to  know  the  product 
of  1.413  multiplied  by  .647.  This  would  be  the  same  as 
wishing  to  know  what  number  bears  the  same  ratio  to 
1.413  that  .647  bears  to  unity. 
Draw  then  two  lines,  AD  and 
A  E,  as  long  as  you  please,  and 
making  what  angle  you  please. 
Open  your  compasses  to  the 
length  of  a  unit,  and  putting  one  foot  at  A  with  the  other 
make  a  dot  at  C.  Open  then  again  to  the  width  .647  (that 
is  to  say,  in  Fig.,  Art.  171,  until  they  measure  the  distance 
from  a  point  in  the  line  6,  7,  seven  tenths  of  the  way  from 
4,  4,  towards  5,  5,  to  the  line  A  B),  and  set  that  off  from 
A  to  B.  Join  the  points  C  and  B  by  a  straight  line.  In 
like  manner  open  your  compasses  to  the  width  1.413,  and 
set  it  off  from  A  to  E.  Draw  E  D  parallel  to  C  B,  and  as 
the  triangles  ABC  and  A  D  E  are  similar,  it  is  manifest 
that  A  D  will  be  the  required  product.  Open  your  com- 
passes till  they  just  stretch  from  A  to  D,  and  you  will  find, 


POSTULATES.  fe9 

on  applying  thorn  to  your  scale,  that  the  lengtli  is  .91G  — 
the  ])rodiicX  of  1.413  by  .648. 

This  would  be  multiplying  two  numbers  by  geometrical 
construction.  You  would  represent  the  numbers  by  lines, 
and  unity  by  a  line,  and  the  product  of  the  numbers  will 
then  be  represented  by  a  line  bearing  the  same  ratio  to  the 
multiplicand  that  the  multiplier  bears  to  unity.  This  line 
is  found  by  drawing  two  similar  triangles ;  and  the  line 
being  measured  gives  the  product  in  figures. 

173.  In  the  simple  illustration  I  have  given,  there  would 
be  no  advantage  in  a  geometrical  construction  over  the 
arithmetical  process.  But  it  is  by  no  means  always  so. 
On  the  contrary,  there  are  a  vast  variety  of  cases  in  which 
geometrical  construction  is  by  far  the  best  method  of  solv- 
ing practical  questions  of  mathematics.  For  this  reason,  I 
recommend  the  scholar  to  make  himself  famihar  with  its 
processes.  The  whole  of  the  second  part  of  this  volume 
is  intended  to  assist  him  in  gaining  this  knowledge. 


CHAPTER    II. 

POSTULATES. 

174.  In  the  same  manner  that  there  are  certain  truths 
too  plain  to  need  proof,  which  are  called  axioms,  so  there 
are  processes  of  mechanical  construction  too  simple  and 
easy  to  make  it  doubtful  whether  they  can  be  performed. 
These  are  sometimes  called  postulates,  that  is,  things 
asked,  because  you  are  asked  to  take  it  for  granted  that 
they  can  be  done. 

175.  Postulate.  It  is  possible  to  have  a  plane  sheet  of 
paper,  —  Of  course  no  surface  of  paper  can  be  perfectly 
plane ;  but  it  is  easy  to  obtain  a  table  so  nearly  plane,  and 


60  POSTULATES. 

paper  so  nearly  plane  in  its  surface,  that  no  appreciable 
eri'or  can  arise  in  using  them  as  plane. 

176.  Postulate.  A  straight  line  can  he  drawn  from  one 
point  to  another  on  plane  paper,  —  This  again  cannot,  of 
course,  be  done  with  perfect  exactness;  but  when  the 
points  are  marked  by  fine  dots,  we  can  bring  the  straight 
edge  of  a  ruler  up  to  the  two  points,  and  draw  a  line  so 
nearly  through  them,  and  so  nearly  straight,  that  there 
shall  be  no  appreciable  error. 

177.  Postulate,  Any  straight  line  may  he  continued 
at  either  end  for  any  distance,  —  This  requires,  of  course, 
the  same  limitation  as  to  accuracy  as  the  preceding  postu- 
late. The  edge  of  the  ruler  is  to  be  applied  to  the  line 
already  drawn,  as  a  guide  in  prolonging  it  farther. 

178.  Postulate,  Around  any  point  as  a  centre  a  circle 
may  he  drawn  of  any  radius,  required,  —  This  postulate, 
also,  must  be  limited  to  mean,  that  this  can  be  done  with- 
out appreciable  error.  But  no  man  can  put  down  one  foot 
of  the  compasses  exactly  in  the  centre  of  a  dot,  nor  open  a 
pair  exactly  to  the  length  of  a  given  line. 

179.  In  the  writings  of  geometers,  usually  these  postu- 
lates are  not  limited,  and  are  used  only  as  the  foundation 
of  theoretical  solutions.  But  as  I  do  not  see  the  value  of 
putting  a  theorem  into  the  form  of  a  problem,  unless  for 
practical  use  in  geometrical  constructions,  I  have  added 
these  limitations. 


STRAIGHT   LINES   AND    ANGLES. 


61 


CHAPTER   III. 


STRAIGHT    LINES    AND    ANGLES. 


180.  Problem,  To  divide  a  line  into  equal  parts,  —  If 
the  parts  are  to  be  two  in  num- 
ber, draw  from  the  ends  of  the 
line,  as  centres,  arcs  o(  equal  ra- 
dii, the  radius  being  long  enough 
to  cause  the  arcs  to  intersect 
each  other  at  two  points.  Join 
the  points  of  intersection  by  a 
straight  line,  and  this  line  will 
intersect  the  given  line  in  the 
middle.  For,  since  B  C  and  A  C  are  equal,  the  angles 
C  A  E  and  C  B  E  are  (Ex.  XI.)  equal ;  and  since  the  tri- 
angles D  A  C  and  D  B  C  are  composed  of  sides  of  the  same 
length,  the  angles  B  C  E  and  ACE  are  (by  Art.  95)  equal. 
Hence,  by  Art.  91,  the  triangle  B  E  C  and  A  E  C  are  equal, 

and  AErr::EB. 

If  the  parts  are  to  be  more  in  number  than  two,  other 
methods  of  division  may  readily  be  devised,  which  shall 
only  require  the  postulates  of  Chap.  II. 

But  a  slight  extension  of  the  postulate  of  Art.  178 
renders  all  methods  of  division  practically  unnecessary. 
The  ability  to  draw  a  circle  around  any  centre  implies 
your  ability  to  put  one  foot  of  the  compasses  on  any  point 
you  choose.  And  the  postulate,  that  you  can  draw  it  with 
any  radius,  implies  that  the  compasses  may  be  opened  to 
any  width  desired.  But  if  we  can  open  the  compasses  to 
extend  from  A  to  B,  it  is  practically  just  as  true  that  we 
can  open  them  to  extend  half  way,  so  that  two  "  steps  " 
shall  take  us  from  A  to  B,  or  one  third,  one  fourth,  &c. ; 
C 


C2  STRAIGHT    LINES    AND    AXGLES. 

the  way,  so  that  three  or  four  steps  may  carry  us  from  one 
end  to  the  other.  If  the  first  effort  does  not  succeed,  tlie 
width  of  the  compasses  is  to  be  altered  in  proportion  to 
the  number  of  steps  we  have  made.  In  dividing  a  line 
into  sevenths,  if  our  seventh  step  left  us  .14  of  an  inch 
from  the  end  of  the  line,  the  compasses  are  to  be  opened 
only  one  fiftieth  of  an  inch  wider  than  before. 

181.  To  divide  any  angle  into  equal  parts.  — A  similar 
extension  of  the  postulate  in  Art.  178  will  show  us  that  we 
can,  without  appreciable  error,  find  a  chord,  which,  being 
applied  a  given  number  of  times  to  a  given  arc,  will  coincide 
at  its  extremities  with  the  extremities  of  the  arc ;  in  other 
words,  we  can  open  the  compasses  to  a  width  Avhicli  will 
step  over  an  arc  in  a  given  number  of  steps. 

Let,  then,  B  A  C  be  the  angle 
to  be  divided.  With  any  radius, 
taking  A  as  the  centre,  draw  an 
arc  B  C  between  the  sides  of  the 
angle.  Step,  with  the  compass- 
es, over  this  arc  in  as  many  ste23S  as  the  parts  into  wliich 
the  angle  is  to  be  divided.  Let  d  and  e  be  two  points  of 
division  thus  determined  nearest  B.  Draw  c^?  A  and  <?  A, 
and  the  angles  eKd  and  c? A B  are  two  of  the  required 
parts  of  the  angle  BAG.  For  if  we  imagine  chords  e  d  and 
c?B  to  be  drawn,  then  the  triangles  eKd  and  c?  A  B  will 
be  composed  of  sides  of  the  same  length,  that  is,  equilat- 
eral with  respect  to  each  other,  and  of  course  equal  in  all 
Irheir  i3arts  (Art.  95)  ;  whence  the  angles  eKd  and  ^?  A B 
will  be  equal.        /■-' 

182.  Corollary.  If  the  figure  eKd  were  tunied  over 
on  the  line  A  J  as  a  hinge,  the  line  A  e  would  coincide 
wkh  A  B,  and,  as  all  parts  of  the  arcs  are  at  equal  distances 
from  A,  the  arcs  ed  and  d^  woi>ld  coincide.  That  is, 
equal  chords  in  the  same  circle  subtend  equal  arcs,  and 
arcs  can  be  equally  divided  by  a  pair  of  dividers,  in  tlie 
same  manner  as  straight  lines  and  angles. 


STRAIGHT   LINES    AND    ANGLES.  63 

183.  Frohlem,  To  draio  an  angle  of  a  given  number 
of  degrees,  —  From  any  point 

A,  as  centre,  in  a  straight  line     ;         r--.,^^  ^ 
A  B,  with  any  radius  A  B,  dc-     j  /'^n 

scribe  an  arc  B  C.     Keeping     •  A^ 

the  compasses  open  at  the  same     j  \ 

width,  place  one  foot  at  B,  and     j  J^D 

with  the  other  mark  the  point     I  ^.^^^^^  \ 

C.     The  arc  B  C  is  then  (Ex.     j         ^^ 

XII.)  an  arc  of  60^     Closing    \k^^^ \^ 

the  compasses  until  they  will 

pass  over  the  same  arc  in  four  steps,  you  obtain  (by  Art. 
182)  arcs  of  15°.  Selecting  either  of  these,  according  to 
the  degrees  required,  close  the  compasses,  until  they  divide 
it  into  arcs  of  5°.  By  dividing  one  of  these  arcs  of  5°  into 
^\Q  equal  parts,  you  can  obtain  the  required  degree,  count- 
ing from  B  up  to  D.  Join  D  A,  and  you  manifestly  have 
the  angle  required.  Thus,  if  the  given  number  of  degrees 
were  twenty-seven,  we  should  take  the  second  arc  of  15°, 
the  third  arc  of  5^  in  that  arc  of  15°,  and  the  second  de- 
gree of  those  five. 

184.  The  formation  of  a  protractor.  —  Take  a  piece 
of  hard,  smooth  card,  draw  a  fine,  straight  line,  as  A  B  (see 
fig.  above),  and  with  a  convenient  radius,  say  three  inches, 
draw  the  arc  B  C.  Measure  carefully  the  arc  B  of  60°  by 
having  the  compasses,  while  yet  unaltered  from  the  radius 
with  which  you  drew  the  arc,  step  from  B  to  C.  Divide 
the  arc  as  accurately  as  possible  into  four  equal  arcs  of  15*^ 
each,  and  set  off  two  such  arcs  beyond  C,  so  as  to  make 
the  whole  arc  90°.  Divide  each  arc  of  15°  carefully  into 
three  equal  parts,  which  will  each  be  5°.  Divide  each  also 
into  five  parts,  each  of  which  will  be  3°.  By  stepping  over 
the  whole  arc  with  the  compasses  open  for  three  degrees, 
first  stepping  over  it  lightly  to  make  sure  that  twenty  steps 
will  exactly  make  60°,  and  then  with  a  heavier  step,  so  as 


64  STRAIGHT   LINES   AND   ANGLES. 

to  leave  footprints  ;  repeating  this  heavier  stepping  from 
each  point  of  division  of  the  5°  arcs,  you  can  divide  t\ie 
prolonged  arc  into  90  equal  degrees.  The  first  divisions, 
starting  from  B,  will  give  3,  6,  9,  12, 15,  18,  21,  &c.  The 
second,  starting  from  the  5°  point,  will  give  2,  6,  8,  11,  14, 
17,  20,  &G,  The  third,  starting  from  the  10°  point,  will 
give  1,  4,  7,  10,  13,  16,  19,  22,  &c. ;  and  these  three  series 
evidently  embrace  all  numbers.  Mark  each  fifth  point 
with  a  longer  mark,  and  number  them  from  B  towards  C. 

185.  The  graduated  arc  and  its  centre,  described  in  Art. 
183,  is  called  a  protractor,  and  may  be  found  for  sale,  en- 
graved on  wood,  ivory,  or  brass.  It  is  used  for  measuring 
angles,  and  also  for  drawing  angles  of  a  given  size.  There 
are  two  ways  in  which  it  can  be  finished  and  used.  The 
first  way  of  measuring  an  angle,  is  to  draw  an  arc  between 
its  sides,  prolonged  if  necessary,  with  the  same  radius  as 
that  of  your  protractor,  its  centre  being  exactly  at  the 
vertex.  Set  the  compasses  so  as  to  reach  exactly  across 
this  arc  from  side  to  side  of  the  angle ;  then,  placing  one 
foot  of  the  compasses  at  the  point  B  of  the  protractor,  the 
other  will  mark  out  on  the  graduated  arc  the  size  of  the 
angle.  The  reverse  process  of  drawing  a  given  angle 
consists  in  drawing  an  arc  of  the  same  radius  as  that  of  the 
protractor,  and  then  with  the  compasses  taking  the  chord 
of  the  given  number  of  degrees  from  the  protractor  and 
setting  it  upon  the  arc ;  lines  drawn  from  these  two  points 
of  the  arc  to  the  centre  from  which  it  was  drawn,  will 
make  the  required  angle. 

The  second  method  of  using  the  protractor  is  to  cut  off 
all  the  card  below  the  line  A  B  and  all  outside  the  gradu- 
ated  arc  B  C.  Placing  then  the  point  A  over  the  vertex 
of  the  angle,  and  making  A  B  coincide  with  one  side  of  the 
angle,  the  other  side  prolonged  if  necessary, -will  P^^s  out 
under  the  graduated  edge  of  the  card,  and  the  degree  of 
the  angle  can  be  at  once  read.    Or,  if  you  wish  to  draw  an 


STRAIGHT   LINES    AND    ANGLES.  65 

angle  of  a  given  size,  having  placed  the  edge  A  B  as  just 
directed,  make  a  dot  on  the  paper,  at  the  right  degree  on 
the  graduated  edge,  and  then,  removing  the  protractor, 
join  the  dot  by  a  straight  line  with  the  vertex  that  was 
under  A. 

Protractors  may  be  purchased  having  graduated  arcs 
of  180°,  or  of  360°.  In  the  latter  case,  the  central  part  of 
the  plate  is  removed,  and  a  piece  of  transparent  mica  in- 
serted, with  a  fine  dot  upon  it  to  mark  the  exact  centre. 

186.  To  draw  an  angle  equal  to  a  given  angle.  —  If 
the  given  angle  is  given  in  degrees,  the  required  angle  mjiy 
be  drawn  by  Art.  185.  But  if  the  given  angle  is  one  sim- 
ply drawn  on  paper,  as  A  B  C,  then  from  the  vertex  of  the 
given  angle  as  a  centre,  with  any  radius,  draw  an  arc  be- 
tween the  sides  of  the  angle ;  and  with  the  vertex  of  the 
required  angle  as  a  centre,  with  the  same  radius,  describe 
an  arc  of  equal  length.  (Art.  182.)  Lines  drawn  through 
the  extremities  of  this  arc  to  the  vertex  will  make  the 
required  angle. 

Thus,  if  it  be  required  to  draw  a  line  from  the  point  A, 
making  the  angle  E  with  the 
line  A  B,  draw  with  any  radius 
the  arc  F  G,  and  with  the  same 
radius  the  arc  C  D,  making  C  D 
equal  to  F  G.      A  line   drawn 

through  the  points  A  and  D  will   ^ ^ -^ 

make  the  required  angle. 

187.  To  draw  tkrougJi  a  given  pointy  as  C,  a  line  paral- 
lel to  a  given  line^  as  A  B.  —  Join  C  to  any  point  in  the 
given  line  by  a  straight 
line,  as  C  A.  Make  the 
angle  D  C  A  equal  to 
CAB,  and  the  line  DC 
will  be  manifestly  par- 
allel to  A  B.    To  draw  through  C  a  line  making  any  angle 

6* 


D                             C 

E 

.-^  y^" 

\ 

66  STRAIGHT   LINES   AND   ANGLES. 

with  A  B,  we  need  only  draw,  from  any  point  in  A  B,  a 
line  making  the  required  angle  with  A  B,  and  then  draw 
through  C  a  line  parallel  with  the  line  so  drawn. 

188.  A  simpler  mode  of  doing  the  same  thing,  though 
not  allowed  by  the  postulates  of  Chaj).  II.,  is  to  open 
the  compasses  until,  with  one  foot  on  the  point  C,  the  other 
will  describe  an  arc  touching  the  line  A  B,  but  not  cutting 
it.  With  the  same  radius  and  one  foot  at  B,  describe  an- 
other arc  at  E.  Draw  a  line  through  C,  touching,  but  not 
cutting,  the  arc  E,  and  it  will  be  parallel  to  A  B.  The 
proof  may  be  readily  discovered  by  the  learner. 

189.  The  instrument  called  a  parallel  ruler  is  simply 

two  rulers  with  par-         i ; 1 . 

allel   edges,  joined         ^A^V — ' ^^*Nr 

by  two    strips    of    . XL _^ 


\5~ 


brass,  riveted  to  the    " '■ 

rulers,  but  the  rivets  allowing  motion  in  the  plane  of  the 
paper  on  which  it  is  laid.  Great  care  must  be  taken  to 
have  the  rivet  holes  in  the  two  pieces  of  brass  at  equal 
distances,  and  also  those  in  the  rulers  at  equal  distances. 
If  this  is  done,  then,  while  one  ruler  is  held  still  and  the 
other  moved,  the  moving  ruler  must  remain  parallel  to  its 
first  position. 

Additional  care  is  usually  taken,  in  making  the  instni- 
ment,  that  this  position  shall  be  parallel  to  that  of  the  sta- 
tionary ruler,  by  having  the  holes  in  each  ruler  on  a  line 
parallel  with  its  edges.  Another  kind  of  parallel  ruler  is 
made  by  simply  mounting  a  ruler  on  rollers.  This  is  less 
accurate. 

The  readiest  and  most  accurate  mode  of  drawing  paral- 
lel lines  is,  however,  to  use  a  flat  triangle  of  wood,  one 
side  of  wliich  is  slid  against  the  edge  of  a  straight  ruler, 
held  firmly  stationary,  while  the  other  sides  remain  paral- 
lel to  their  first  position. 

190.  The  drawing  of  a  parallel  line  is  simply  the  draw- 


STRAIGHT   LIXES   AND    ANGLES.  67 

ing  of  an  angle  equal  to  zero.  Another  angle  of  pecnliar 
interest  is  the  right  angle,  and  there  are  better  ways  than 
that  of  Art.  185  for  drawing  a  right  angle. 

191.  2'o  raise  a  perpefidicular  at  a  given  point  A 
upon  a  straight  line  A 13.  —  I^irst  Method,  From  any 
point  B,  in  the  line  A  B,  with  Dn 
any  radius  B  C,  describe  an 
arc,  and  from  the  point  A, 
with  the  same  radius,  describe 
an  arc  cutting  the  first  at  C. 
Draw  the  line  B  C,  prolonging 
C  D  to  equal  C  B.  Join  D  A 
by  a  straight  line,  and  it  will 
be  perpendicular  to  A  B,  by  Ait.  147. 

Second  Method,  Opening  the  compasses  to  any  con- 
venient width,  step  off  ^nq  equal  portions  of  A  B,  begin- 
ning at  A.  Let  B  be  the  fourth  point  of  division.  From 
B  as  a  centre,  with  a  radius  equal  to  five  of  these  parts, 
(^•aw  an  arc  above  A,  and  from  A  as  a  centre,  with  a  ra- 
dius equal  to  three  parts,  draw  a  second  arc  intersecting 
the  first  at  D.  Join  D  to  A  by  a  straight  line,  and  it  will 
be  at  right  angles  to  A  B,  by  Art.  106. 

Third  Method,  Measure  with  the  compasses  equal  dis- 
tances on  each  side  of  A,  and  bisect  the  line  thus  meas- 
ured off  by  the  method  of  Art.  180,  and  you  have  a  line 
passing  through  A  at  right  angles  to  A  B. 

Fourth  Method,  Visiting  or  business  cards  are  usually 
cut  very  exactly  at  right  angles.  By  applying  one  corner 
of  a  card  at  A,  and  making  one  edge  coincide  with  A  B, 
the  other  edge  will  be  at  right  angles  to  A  B.  The  accu- 
racy of  this  right  angle  may  be  tested  by  drawing  perpen- 
diculars on  opposite  sides  of  A. 

Fifth  Method,  If  you  have  neither  card  nor  compasses, 
fold  a  piece  of  writing  paper  carefully,  and  then  double  the 
folded  edge  carefully  on  itself.    This  corner  of  four  thick- 


68 


STRA.IGHT   LINES   AND    ANGLES. 


nesses  of  paper  will  be  a  square  corner,  to  be  used  as  the 
card. 

Sixth  Method.  From  any  point  outside  the  line,  as  C, 
with  a  radius  equal  to  C  A,  draw  an  arc  cutting  A  B,  say- 
in  B,  and  prolonged  to  a  point  where  the  radius  B  C,  pro- 
longed through  C,  may  cut  it,  say  in  D.  DA  will  then, 
by  Art.  147,  be  the  perpendicular  required. 

192.  To  let  fall  a  perpendicular  from  a  pointy  as  D, 
upon  a  straight  line,  as  A  B.  D^ 
—  I^irst  Method.  Join  any 
point  of  the  line  A  B,  as,  for  in- 
stance, the  point  B,  to  the  point 
D,  by  a  straight  line  D  B.  From 
C,  the  middle  of  D  B,  with  a 
radius  equal  to  CB  or  CD, 
draw  an  arc  cutting  A  B  at  E, 
and  D  E  will  be  (by  Art.  147)  the  peipendicular  required. 

/Second  Method.    From  D  as  a  p 

centre,  describe  any  arc  cutting 
the  line  A  B  in  two  places.  The 
middle  point  between  these  places 
will  be  E,  the  foot  of  the  perpen- 
dicular from  D. 

Third  Method.  Make  one  side  of  the  square  card  coin- 
cide with  A  B,  and  slip  the  card  along  until  the  end  passes 
through  D.  Or  if  the  triangle  of  Art.  189  be  made  with  an- 
gles of  30\  60°,  and  90°,  its  square  corner  may  be  used. 

Fourth  Method.  If  the  perpendicular  is  to  be  drawn 
merely  for  the  sake  of  measuring  its  length,  that  is,  for 
finding  the  distance  of  D  from  the  line  A  B,  it  need  not 
be  drawn  ;  but  you  may  simply  place  one  foot  of  the  com- 
I^asses  in  D,  and  then  open  them  wide  enough  to  desciibe 
an  arc  touching,  but  not  cutting,  A  B.  Although  the  pos- 
sibility of  doing  this  is  not  claimed  in  the  postulates,  yet  it 
is  practically  equivalent  to  the  postulate  of  Art.  178. 


TKIANGLES.  69 

CHAPTER   ly. 

TRIANGLES. 

195.  To  draio  a  triaiigle  of  three  given  sides,  —  If  the 
eides  are  not  drawn,  but  are  given  in  numbers,  open  the 
compasses  to  extend  upon  the  scale  to  a  number  corre- 
sponding to  one  of  the  sides.  If  the  lines  are  drawn,  open 
the  compasses  to  the  length  of  one  of  them.  Set  the 
dividers  on  paper  with  sufficient  pressure  to  mark  the 
points  where  the  feet  touch.  From  these  points  as  cen- 
tres, with  radii  equal  to  the  other  sides  of  the  triangle, 
draw  arcs  cutting  each  other.  This  point  of  intersection 
and  the  points  used  as  centres  will  be  the  vertices  of 
the  required  triangle,  and  must  be  joined  by  straight 
lines. 

194.  To  draw  a  triangle  xolien  tioo  sides  and  one  angle 
are  given.  —  First  Case.  When  the  angle  is  included 
heticeen  the  given  sides.  Draw  two  lines  making  the  re- 
quired angle ;  and  upon  each  line  set  off  with  the  com- 
passes, from  the  vertex,  the  length  of  the  given  sides ;  join 
their  extremities  by  a  straight  line,  and  you  evidently  have 
the  required  triangle. 

Second  Case.     When  the  angle  is  opposite  one  of  the 
given  sides.    Draw  two  lines,  A  B 
and  A  C,  making  the  given  angle. 
Set  off  from  the  vertex  A  the  given 
adjacent  side  A  B,  and  from  B  as  a 
centre,  with   the   other  given   side 
B  C  as  a  radius,  draw  an  arc  cutting  A  C  in  C  or  c.    Ei- 
ther A  B  C  or  A  B  c  will  be  the  required  triangle.     If  B  C 
is  greater  than  A  B,  only  on<i  triangle  can  be  formed. 

195.  To  draw  a  triangle  when  09ie  side  and  two  angles 


70  TRIANGLES. 

are  gimn,  — First  Case,  When  the  side  lies' between  the 
angles.  Draw  a  line  equal  in  length  to  the  given  side, 
and  draw  at  the  ends  of  it  lines  making  the  given  angles 
with  the  given  side.  These  lines  being  produced  far 
enough  to  meet,  will  give  the  required  triangle. 

Second  Case.  W7ien  one  angle  is  opposite  the  given 
side.  If  the  angles  are  given  in  degrees,  the  simplest  way 
is  to  add  the  two  angles  together,  and  subtract  the  sum 
from  180' ;  this  will  give  the  third  angle,  and  reduce  this 
case  to  the  first  case.  But  if  the  angles  are  given  by- 
being  drawn,  it  will  be  better  to  draw  the  given  side  A  B, 
and  at  one  end  raise  the  line  A  C,  mak-  ^   C 

ing  the  given  adjacent  angle.  At  any 
point,  as  C,  draw  C  D,  making  A  C  D 
equal  to  the  given  opposite  angle. 
Through  B  draw  B  E  parallel  to  CD,  and  B  E  A  is  evi- 
dently the  required  triangle.  Such  a  line  as  C  D  should 
be  drawn  lightly,  so  that,  if  necessary,  it  can  be  erased. 

196.  It  is  manifest  that,  in  all  the  problems  of  this 
chapter,  if  the  sides  are  given  in  numbers,  any  convenient 
unit  may  be  taken  to  represent  unity  in  the  numbers. 
That  is  to  say,  if  the  original  numbers  represent  feet, 
yards,  or  miles,  they  may  in  your  drawing  be  taken  as 
inches,  tenths  of  inches,  twentieths,  or  hundredths,  as  you 
please ;  only  remembering  that  the  same  quantity  must 
be  taken  as  the  unit  in  all  parts  of  any  one  figure. 

In  drawing  profiles,  or  vertical  sections,  however,  two 
units  are  usually  employed.  Thus,  in  drawing  a  sketch 
of  the  elevations  and  depressions  of  a  railroad  100  miles 
long,  in  which  the  greatest  elevation  attained  was  500  feet, 
you  might  represent  the  length  on  a  scale  of  one  mile  to 
an  inch,  but  the  elevations  and  depressions  on  a  scale  of 
400  feet  or  500  feet  to  an  inch. 

197.  The  problem  of  Art.  193  is  impossible  if  either  of 
the  given  sides  is  greater  than  the  sum  of  the  other  two. 


TRIANGLES.  71 

198.  In  the  second  case  of  Art.  194,  the  problem  is  im- 
possible if  the  side  opposite  the  given  angle  is  too  short 
to  reach  the  side  not  given. 

199.  In  Art.  195,  the  problem  is  impossible  if  the  sum 
of  the  angles  given  equals  or  exceeds  180°. 

Examples. 

200.  Draw  a  triangle,  with  sides  of  19,  33,  and  41  feet. 
Draw  one  with  sides  of  41,  33,  and  19  miles.  Draw  trian- 
gles with  sides  of  18,  13,  and  27  ;  of  341,  263, 501 ;  of  76.8, 
54,  43.7  ;  of  673,  321,  352;  of  67,  32,  34;  of  71,  39,  43  ;  of 
67,  29,  47. 

201.  Measure,  by  Art.  185,  the  angles  of  each  triangle 
in  Art.  200,  and  test  the  accuracy  of  your  measures  by 
adding  the  angles  of  each  triangle  together;  the  sum 
should,  of  course,  be  180°. 

202.  Draw  a  map  of  a  triangular  building  lot,  whose 
sides  are  97  and  73  feet,  the  angle  between  these  sides 
being  57"^.  Draw  a  map  of  another  triangular  lot,  with 
sides  of  84  and  77  feet,  and  the  angle  opposite  the  side  of 
77  feet  equal  to  43°.  Draw  a  triangle  with  a  side  of  81, 
another  of  41,  and  the  angle  opposite  41  equal  to  23°. 
Try  the  same  with  the  angle  37°. 

203.  Measure  the  other  angles  of  the  triangles  of  Art. 
202,  and  the  third  sides,  testing  the  angles  as  in  Art.  201. 

204.  One  side  of  a  triangular  lot  being  83,  what  is  the 
size  of  the  opposite  angle,  and  the  length  of  the  other  two 
sides,  the  adjacent  angles  being  67°  and  93""-  Answer  the 
same  question  when  the  angles  are  33°  and  111°.  When 
the  angles  are  61°  and  119°. 

205.  If  we  have  only  the  angles  of  a  triangle  given,  we 
of  course  cannot  discover  the  length  of  the  sides. 

206.  It  is  plain  that,  having  two  sides  given,  we  must 
also  have  an  angle  given,  in  order  to  draw  the  triangle. 


72  QUADRANGLES. 

207.  If  one  side  is  given,  it  is  plain  that  we  must  have 
two  angles  given,  or,  having  but  one  actually  given,  must 
have  some  condition  given  which  will  determine  another ; 
such,  for  instance,  as  the  ratio  which  the  unknown  angles 
bear  to  each  other. 

208.  A  ship  at  anchor  finds  that  a  round  lighthouse- 
tower,  known  to  be  17  feet  in  diameter,  covers  a  degree 
and  a  half  of  the  horizon  ;  in  other  words,  lines  drawn  from 
the  ship  to  oj^posite  sides  of  the  tower  make  an  angle  of 
1°  30' ;  and  those  lines  make,  with  one  diameter  of  the 
tower,  equal  angles  of  89°  15'  each.  Draw  this  triangle, 
and  find  from  it  the  distance  of  the  ship  from  the  light- 
house. 

What  is  the  moon's  diameter,  if  her  distance  from  the 
earth  is  240,000  miles,  and  her  apparent  diameter  30'  ? 


CHAPTER   V. 

QUADRANGLES. 

209.  To  draw  a  quadrangle  when  all  the  sides  and  one ' 
angle  are  given;   the  sides,  i7icluding   the  angle^  being 
named, — Draw  two  lines,   making    the 
given  angle.  A,  and  measure  upon  them 
the  sides,  including  the  angle.     From  the 
extremities,  B,  C,  of  these  sides,  as  cen- 
tres, with  radii  equal  to  the  other  sides, 
draw  arcs  intersecting  each  other,  and  the 
point   of  intersection   will  be   the  fourth  vertex   of  the 
quadrangle. 

210.  It  is  manifest  that  the  arcs  will  intersect  in  two 
places,  and  also  that  the  third  and  fourth  sides  can  change 


QUADRANGLES. 


73 


places  with  each  other,  so  that  four  quadrangles  can  some- 
times be  drawn  satisfying  the  conditions  of  Art.  209. 

211.  To  draw  a  quadrangle  when  three  sides  and  two 
angles  are  given^  the  order  of  the  sides  and  angles  being 
named,  —  If  both  the  angles  are  included  between  given 
sides,  draw  the  middle  side  and  raise  the  other  two  sides, 
making  the  proper  angles  with  the  middle  side.  It  will 
only  remain  to  join  their  extremities  by  a  straight  line. 

If  one  angle  is  adjacent  to  the  unknown  side,  as,  for 

instance,  if  AB,  A  D,  and  D  c  are  given  A B 

sides,  and  A  and  B  given  angles,  draw 
A  B  and  AD  of  the  given  lengths  and 
making  the  given  angle.  From  B  draw 
B  C  of  indefinite  length,  making  B  of  the 
given  size.  From  D  as  a  centre,  with  the 
given  length  D  C  as  a  radius,  describe  an 
arc  cutting  B  C  in  C  and  c.  Join  either 
D  C  or  D  c,  and  it  completes  the  quadrangle. 

But  if  both  given  angles  are  adjacent  to  the  unknown 
side,  that  is, suppose  Be  the  unknown 
side,  and  B  and  C  the  given  angles, 
then  at  B,  and  at  any  other  point  on 
B  c,  as  c,  raise  at  the  proper  angles  the 
sides  B  A  and  c  d  of  the  given  length. 
From  the  point  A,  with  the  radius  A  D, 
draw  an  arc.  From  d  draw,  parallel 
to  c  B,  a  line  cutting  this  arc  at  D. 
Draw  D  C  parallel  to  d  c,  and  the 
quadrangle  is  manifestly  completed. 

212.  To  draw  a  quadrangle  ichen  two  sides  and  three 
angles  are  given^  their  order  of  position  being  named,  — 
The  three  angles  being  subtracted  from  360°  will  give  the 
fourth  angle. 

If  the  given  sides  are  adjacent,  as  A  B,  A  D,  draw  those 
sides  of  the  given  length,  making  the  given  angle,  and 
7 


7^  CIRCLES. 

from  the  extremities,  B  and  D,  draw  lines  making  the 
given  angles  B  and  D.  The  intersection  of  these  lines 
will  complete  the  quadrangle. 

If  the  given  sides  are  not  adjacent,  as  A  B  and  D  C, 
from  the  extremities  of  A  B,  draw  lines  A  D  and  B  C,  of 
any  length,  but  making  the  given  angles  A  and  B.  From 
any  point  c  in  B  c  draw  c  c7,  making  the  angle  Bed  equal 
to  the  given  angle  C.  Take  c  d  equal  to  the  given  side 
D  c.  Draw  c?D  parallel  to  c  B.  The  point  at  which  dY> 
crosses  A  D  will  be  one  corner  of  the  quadrangle.  From 
this  point  D  draw  D  C  parallel  to  d  c,  and  the  quadrangle 
is  manifestly  completed. 

213.  My  garden  is  an  iiTegular  quadrangle,  the  sides 
being  150,  207,  315,  and  97  feet.  The  sides  are  placed  in 
that  order,  and  the  angle  between  the  first  and  second  is 
96°.     Draw  me  a  plot. 

The  south  front  of  a  lot  is  37  feet,  the  east  side  63,  the 
west  52  feet,  and  the  south-west  corner  is  92'',  the  south- 
east 62°.    Draw  a  map. 

The  south  front  of  a  lot  of  land  being  31  feet,  the  east 
side  53,  the  west  46  feet,  the  south-west  corner  is  a  right 
angle,  the  north-east  corner  measures  78°.     Draw  a  map. 

The  sides  of  another  lot,  and  its  south  front,  measure  \\\q 
same  as  in  the  last  example,  but  the  two  corners  in  the 
rear  are  square  corners.    Draw  a  map. 


CHAPTER   VI, 

CIRCLES. 


214.  To  dram  an  arc  of  a  circle^  the  radius  being  given, 
—  As  a  practical  question  this  is  one  of  great  importance, 
as  it  concerns  not  only  the  process  of  geometrical  cofl^- 


CIECLES.  75 

struction,  but  many  processes  of  mechanical  construction 
also.  In  geometrical  construction  the  compasses  usually 
afford  the  readiest  means  of  draAving  arcs  and  circumfer- 
ences ;  but  in  mechanical  construction  of  machines,  roads, 
and  other  things,  the  compasses  are  frequently  of  no 
value. 

Sometimes  a  thread,  string,  or  rope  is  fastened  by  one 
end  to  the  spot  selected  for  the  centre  of  the  arc,  while 
the  other  end  is  carried  round.  It  is  plain  thataf  the  line 
is  kept  stretched,  and  equally  stretched,  its  moving  end 
remains  always  at  the  same  distance  from  the  station- 
ary end,  and  the  curve  must  be  the  circumference  of  a 
circle. 

Sometimes  the  thing  on  which  you  wish  to  describe  a 
circle  is  turned  round,  as  in  a  lathe. 

Sometimes,  at  the  blacksmith's  shop,  a  circle  is  made  by 
bending  a  strip  of  iron  equally  at  every  part.  This  is  done 
by  passing  it  between  three  rollers.  The  amount  of  bend- 
ing at  each  point  must  be  inconceivably  small,  because  any 
perceptible  bend  at  any  one  point  would  make  an  angle 
there.  Yet  small  as  the  bending  at  each  point  must  be,  it 
must  be  practically  measured  by  a  screw  that  raiseg  or  de- 
presses the  middle  roller,  and  thus  alters  the  curvature  of 
the  tire  as  it  passes  between  them. 

In  laying  out  railroads,  arcs  of  circles  are  drawn  by 
measuring  off  equal  angles 
from  one  point,  as  A,  and 
setting  off  equal  chords  be- 
tween the  sides  of  the  angles, 
beginning  at  the  point  A. 
That  these  will  be  chords  to 
a  circle  may  be  proved  from 
Art.  145.  Engineers  have 
tables  prepared,  telling  them 
what  the  angle  must  be  in  order  to  have  chords  of  100 


76  CIRCLES. 

feet  each  from  circles  of  300  feet,  400  feet,  or  any  other 
radius. 

In  laying  out  garden  paths  and  walks,  it  is  convenient 
to  have  a  wooden  "  square,"  such  as  A  B  D.  A  B  should  be 
straight,  and  have  a  -q 

mark  in  the  middle  ^  c  E 

at  C.    BD  should  ' 


be  divided  into  small  divisions.  Putting  down  two  stakes, 
one  at  A  afid  one  at  C,  a  third  one  may  be  placed  on  D, 
at  such  a  distance  from  B  as  is  desirable.  Taking  up  the 
square,  now  place  the  end  A  at  the  stake  which  was  at  C, 
and  place  the  point  C  at  the  stake  which  was  at  D.  Put 
down  a  fourth  stake  on  the  side  D  as  far  from  B  as  the 
third  stake  was  placed.  The  size  of  the  circle  will  depend 
upon  the  distance  from  B  at  which  the  stake  on  the  side 
D  is  placed.  For  a  small  circle,  divide  C  B  in  the  middle 
at  E,  and  use  the  instrument  as  though  it  had  been  cut  off 
at  C.  By  using  the  whole  for  a  small  circle,  D  is  carried 
too  far  from  B,  and  thus  C  D  becomes  longer  than  A  C, 
which  will  make  the  first  few  stakes  irregular. 

215.  I^irst  Solution  of  Art  214.  Open  the  compasses 
to  the  given  radius,  and  draw  the  arc  as  usual. 

216.  Second  Solution,  If  the  radius  is  too  long  for  the 
compasses  to  be  opened  to  that  width,  we  may  use  a  string, 
a  strip  of  wood  or  of  paper.  If  a  pin  be  thrust  through 
one  end  of  a  strip  of  stiff  paper,  and  a  pencil  point  be  in- 
serted through  a  small  hole  at  the  required  distance  on 
the  strip,  very  accurate  arcs  of  circles  may  be  drawn,  and 
the  radius  measured  beforehand  with  accuracy.  The  pin 
is  held  at  the  centre  of  the  circle,  and  the  pencil  carried 
round. 

217.  Third  Solution.  When  there  is  no  convenient 
place  to  set  the  central  foot  of  the  compasses,  or  the  pin  in 
the  paper  strip,  upon,  other  plans  may  be  adopted. 

Draw  two  lines  on  a  piece  of  stiff  paper  at  an  angle  of 


CIRCLES.  77 

165°  31'.  Trim  the  paper  off  to  these  lines,  leaving  a 
small  piece  about  the  intersec- 
tion. At  the  point  of  intersec- 
tion make  a  small  hole  in  the 
paper  to  insert  a  lead  pencil. 
Placing  two  pins  P  P  at  a 
distance  apart  equal  to  half  the 
radius,  a  neat  arc  may  thus  be  drawn.  Other  angles  may 
be  used  with  corresponding  parts  of  the  radius.  For  in- 
stance, 170°  24',  with  pins  at  a  distance  apart  equal  to  one 
third  the  radius.  Of  course,  in  drawing  these  angles  it  is 
easier  to  measure  the  supplement  of  the  angle,  that  is,  the 
remainder  after  subtracting  it  from  180°,  and  then  simply  , 
let  the  lines  cross. 

218.  Fourth  Solution,  If  we  draw  a  tangent  to  a  cir- 
cle, and  measure  off  upon  it,  from  the  point  of  contact,  dis- 
tances equal  to  one  tenth  the  radius,  we  shall  find  the  dis- 
tance of  the  tangent  to  the  circumference  at  the  first  three 
points  of  division  equal  to  .005,  .020,  .046  of  the  radius. 
Hence  we  may  draw  an  arc  of  a  given  radius  by  drawing 
a  straight  line,  and  marking  upon  it  half  a  dozen  spots 
equally  distant,  at  a  distance  equal  to  one  tenth  the  radius. 
If  over  any  one  of  these  we  make  a  dot  at  the  distance  of 
five  thousandths  of  the  radius,  over  the  next  a  dot  at  the 
distance  of  twenty  thousandths,  &c.,  a  curve  drawn  care- 
fully through  these  dots  will  be  an  arc  as  required.  If  we 
take  twentieths  of  the  radius  instead  of  tenths,  the  distance 
of  the  curve  will  be  .001,  .005,  .011,  .020,  .031,  .046. 

219.  To  draw  through  a  given  point  a  tangent  to  a 
given  circle,  —  First  Case,  If  the  point  is  inside  the  cir- 
cle, the  problem  is  insoluble. 

Second  Case,  If  the  point  is  in  the  circumference,  draw 
aradiustothe  circumference  at  that  point.  Draw  a  line 
at  right  angles  to  the  end  of  the  radius,  and  it  will  be  a 
tangent. 


78  CIRCLES. 

Third  Case.  When  the  point  is  outside  the  circle, 
let  A  be  the  point  and  C  the  cen- 
tre of  the  circle.  Join  the  point 
A  to  the  centre  C  by  a  straight 
line.  From  the  middle  point  B, 
of  that  line,  with  a  radius  B  C, 
equal  to  half  the  line,  draw  an  arc 
which  will  cut  the  circumference  at 
the  points  through  which  the  tan- 
gents from  A  must  pass.  For  it  is 
manifest  that  if  the  angle  ADC  were  drawn,  it  would  be 
a  right  angle,  by  Art.  147. 

Practically^  it  is  only  necessary  to  lay  the  ruler  with  its 
edge  upon  A,  and  touching  the  circumference  without 
cutting.  This  is  not,  however,  practically  useful  when  A 
is  very  nearly  touching  the  circumference,  and  not  at  all 
practical  when  A  is  actually  on  the  circumference,  as  in 
that  case  the  angle  of  the  line  with  the  radius  to  the 
point  of  contact  might  fail  to  be  a  right  angle. 

220.  Draw  a  circle  of  two  inches  radius.  A  semicircle 
of  three  inches  diameter.  Draw,  by  Art.  217,  an  arc  three 
inches  long,  with  a  radius  of  sixteen  inches.  Draw, 
by  Art.  218,  an  arc  five  inches  long,  with  a  radius  of 
ten  inches,  also  of  twenty  inches.  What  angle  does  a 
tangent  through  A  make  with  a  straight  line  drawn  from 
A  to  the  centre  of  the  circle,  the  radius  being  1.5  inches, 
and  the  distance  from  A  to  the  centre  3  inches  ?  Over 
how  many  degrees  of  latitude  could  you  then  look  from  a 
balloon  at  a  height  of  four  thousand  miles  above  the  sea, 
supposing  it  possible  to  rise  that  height?  Over  how  many 
degrees  of  latitude  could  you  look  from  a  balloon  at  the 
height  of  one  thousand  miles  ? 

221.  To  inscribe  a  circle  in  a  triangle,  —  Bisect  two  of 
the  angles  of  the  triangle;  that  is,  divide  each  angle  into 
two  equal  parts  by  a  straight  line.    The  point  where  these 


I 


CIRCLES.  79 

two  lines  intersect  each  other  will  be  the  centre  of  the 

required  circle,  and  the  radius  will 

be   the   length  of  a  perpendicular 

to  either  side  of  the  triangle.   That 

these  perpendiculars  will  be  equal 

in  length,   may  readily  be  shown 

from  the  equality  of  the  triangles 

which  they  form.     Thus  the  triangles  A  F  D  and  A  E  D 

are  equal,  because  they  have  the  same  line  A  D  for  hy- 

pothenuse,  and  equal  angles,  by  the  construction  of  the 

figure.     Therefore  D  F  rz:  D  E.    In  like  manner,  D  G  may 

may  be  shown  to  be  equal  to  D  F. 

222.  To  circumscribe  a  circle  about  a  triangle.  In 
other  words.  To  draio  a  circumference  that  will  pass 
through  three  given  points^  as  the  vertices  of  a  triangle.  — 
As  the  centre  of  the  circle  must  be  equally  distant  from 
each  of  the  points,  it  must  be  found  on  a  line  perpendicu- 
lar to  the  middle  of  a  line  joining  any  two  points.  (Art. 
136.)  In  other  words,  if  we  draw  lines  perpendicular  to 
two  sides  of  the  triangle,  at  the  middle  of  those  sides,  the 
centre  will  be  in  both  these  lines ;  that  is,  will  be  found  at 
their  intersection. 

223.  To  find  the  centre  of  a  given  arc,  —  It  is  manifest 
from  Art.  222,  that  we  only  need  draw  any  two  chords  in 
the  arc,  and  erect  perpendiculars  at  the  middle  of  each 
chord. 

224.  To  find  the  radius  of  a  circumference  that  will  pass 
through  three  given  points  when  they  lie  nearly  in  a 
straight  line.  In  other  words,  WTien  the  radius  is  large^ 
to  draio  the  arc  without  finding  the  centre,  —  The  arc 
may  be  drawn  by  setting  up  pins  at  the  extreme  points, 
and  cutting  a  piece  of  paper  with  straight  edges  at  the 
angle  that  is  made  by  lines  to  the  intermediate  point, 
then  proceeding  as  in  Art.  217.  The  length  of  the  radius 
may  be  found  by  measuring  the  supplement  of  this  angle. 


80  CIRCLES. 

According  as  it  measures  1%  2°,  3^  4°,  5°,  6°,  7^  8%  9°,  or 
10°,  so  must  you  multiply  the  distance  between  the  ex- 
treme points  by  the  number  28.65,  14.33,  9.55,  7.17,  5.74, 
4.78,  4.12,  3.59,  3.20,  or  2.88,  to  obtain  the  radius.  It  will 
be  noticed  how  nearly  any  of  these  numbers  can  be  obtained 
by  dividing  28.65  by  the  number  of  degrees  in  the  supple- 
ment ;  this  will  enable  you  to  obtain  it  for  angles  not  con- 
sisting of  whole  degrees.  Thus  for  the  angle  2J  degr^s, 
you  will  divide  28.65  by  five  halves. 

225.  Make  three  dots  upon  paper  nearly  in  a  straight 
line,  and  discover  by  Art.  224  the  radius  of  a  circle  that 
will  pass  through  them. 

226.  To  describe  an  equilateral  triangle  in  a  circle,  — 
Step  round  the  circumference  with  the  radius,  and  you  will 
require  six  steps  (since  by  joining  two  of  the  points  thus 
marked  with  each  other  and  with  the  centre  of  the  circle 
you  form  an  equilateral  triangle,  whose  angles  must  each 
equal  \  of  180°,  or  \  of  360)  :  by  joining  the  alternate 
points  of  division  with  straight  lines  you  draw  the  triangle 
required.  ^ 

227.  To  describe  a  hexagon  in  a  circle,  —  Step  round  the 
circumference  with  the  radius,  and  join  each  point,  thus 
marked,  with  the  adjacent  points,  by  straight  lines :  the 
hexagon  is  drawn. 

228.  To  describe  a  square  in  a  given  circle.  —  Bisect 
the  arcs  to  which  two  opposite  sides  of  the  hexagon  are 
chords,  join  the  points  of  bisection  with  the  two  vertices 
of  the  hexagon  that  are  at  90° 
from  them,  and  you  will  mani- 
festly have  drawn  a  square. 

229.  To  describe  a  regular  pen- 
tagon or  jive-sided  figure  in  a  cir- 
cle, —  Draw  a  diameter  H  G,  and 
erect  a  radius  C  F  perpendicular  to 
it.     From  D,  at  the  bisection  of  C  G,  measure  D  E  equal  to 


AREAS.  81 

D  F.  Join  E  to  F  by  a  straight  line,  and  it  will  be  equal 
in  length  to  one  side  of  a  pentagon. 

The  proof  of  this  proposition  is  somewhat  intricate,  and 
would  require  more  use  of  algebraic  language  than  is  con- 
sistent with  the  design  of  this  little  book.  Supposing  the 
radius  C  F  to  be  1 ;  then  C  D  would  equal  ^.  From  this 
we  should  find  the  value  of  D  F  by  the  Pythagorean 
proposition.  E  C  would  then  be  found  by  subtracting  ^ 
from  D  F.  Hence,  by  the  Pythagorean  proposition,  we 
could  find  E  F.  Next,  supposing  a  chord  of  72°  (the  side 
of  a  pentagon)  drawn  in  a  circle  of  radius  of  unity,  we 
could  show  that  its  length  would  be  precisely  the  same 
as  that  of  E  F. 

But  the  simplest  way  to  draw  a  pentagon  is  to  open  the 
compasses  to,  as  nearly  as  you  can  estimate  it,  the  fifth  of 
a  circumference,  and  after  stepping  round  once,  alter  their 
width,  as  nearly  as  you  can  estimate  it,  the  fifth  part  of  the 
resulting  error.  This  really  conforms  to  the  spirit  of  the 
postulate,  that  one  can  open  the  compasses  to  a  given 
radius ;  and  the  preceding  method  is  given  simply  to  show 
a  way  of  drawing  a  pentagon  in  conformity  with  the  let- 
ter of  the  postulate.  If  the  compasses  are  opened  so  as 
to  step  round  five  times  without  any  apparent  resulting 
error,  their  real  error  is  probably  but  one  fifth  as  great  as 
it  usually  is  in  measuring  a  radius. 


CHAPTER    VII. 


AREAS. 


230.  Areas  are  the  numbers  which  measure  surfaces; 
that  is,  which  express  the  ratio  of  the  surfaces  to  a  unit 
surface,  or  surface  adopted  as  a  unit,  or  standard  of  ref- 
erence. 


82  AREAS. 

231.  The  usual  unit  of  surface  is  a  square  whose  side  is 
a  linear  unit ;  for  instance,  a  square  inch,  square  mile,  &c. 

232.  To  find  the  area  of  a  rectangle.  —  Multiply  the 
length  of  a  side  by  that  of  an  end,  and  the  product  will 
be  the  area.     (Art.  97.) 

233.  To  find  the  area  of  a  parallelogram,  —  Multiply 
the  length  of  a  side  by  the  distance  to  the  opposite  side ; 
the  product  will  be  the  area.     (Art.  110.) 

234.  To  find  the  area  of  a  triangle,  —  Multiply  the 
length  of  either  side  by  the  distance  from  the  opposite 
vertex;  the  product  will  be  twice  the  area.  (Art.  111.) 

235.  To  find  the  area  of  any  polygon.  —  Divide  the 
polygon  by  diagonals  —  that  is,  by  lines  drawn  through 
vertices  not  adjacent  —  into  triangles,  and  measure  these 
triangles. 

236.  To  multiply  two  numbers  by  geometrical  construc- 
tion. —  This  problem  is  not  commonly  practically  useful, 
and  yet  to  one  who  wishes  only  approximative  results,  and 
dislikes  numerical  computation,  it  may  be  made  to  yield 
good  results,  especially  if  care  be  taken  in  using  the  par- 
allel ruler. 

Have  prepared,  on  a  piece  of  hard,  smooth  paper,  two 
lines  at  an  angle  of  30°  or  40°,  and  on  one  of  them  a  unit, 
AB,  measured  from  the  vertex 
A,  and  permanently  marked.  Lay 
off  either  number  from  A  to  C,  the 
other  from  A  to  D.  Join  B  C, 
and  draw  D  E  parallel  to  B  C. 
A  E  is  the  required  product.  That  is,  if  A  B  were  an 
inch,  A  C  1^  inches,  and  A  D  2  inches,  then  A  E  would  be 
found  to  be  3  inches. 

Proof.  The  triangles  ABC  and  A  D  E  given,  by  Art. 
99  the  proportion  A  E  is  to  A  D  as  A  C  is  to  A  B,  that  is, 
as  A  C  is  to  unity.  But  this  is  the  definition  of  a  product, 
that  it  is  a  quantity  bearing  the  same  ratio  to  the  multipli- 
cand that  the  multiplier  does  to  unity. 


cmcLEs.  83 

287.  To  find  the  area  of  a  circle,  —  If  the  circumfer- 
ence, instead  of  being  a  curve,  consisted  of  many  millions 
of  short,  straight  sides,  it  is  plain  that  the  circle  could  be 
divided  into  many  millions  of  little  triangles,  by  means  of 
many  millions  of  radii.  Now,  the  circumference  may  be 
thus  conceived  of,  and  the  area  of  these  triangles  may  be 
found,  according  to  Art.  234,  by  multiplying  the  circum- 
ference (which  is  the  sum  of  the  short  sides)  by  the  radius 
(the  distance  to  the  opposite  vertex),  and  dividing  by  two. 
The  circumference  or  the  radius  may  be  divided  by  two 
before  they  are  multiplied.  (Art.  157.) 

238.  To  find  the  circumference  of  a  circle  from  knowing 
its  radius.  —  It  is  manifest  from  Arts.  237  and  99  that  the 
ratio  of  the  circumference  to  the  radius  is  the  same  in  all 
circles,  and  we  have  only  to  find  the  circumference  of  the 
circle  whose  radius  is  unity.  Suppose, 
then,  that  B  E  is  a  chord  of  60°,  bi- 
sected at  D,  and  that  A  is  the  centre 
of  the  circle.  Knowing  D  B  is  equal 
to  i,  and  A  B  to  1,  we  can,  by  the 
Pythagorean  proposition,  calculate  the 
length  of  A  D.  Subtracting  this  from 
1  gives  us  D  C.  Then,  knowing  D  C 
and  D  B,  we  can,  by  the  same  proposition,  calculate  B  C. 
Bisecting  B  C  at  c?,  we  know  A  C  and  C  d,  and  can  therefore 
calculate  A  d,  and  thus  find  d  c.  Hence  we  get  the  chord 
C  c,  or  chord  of  ^V  ^^  ^  circle.  By  continuing  this  process 
of  applying  the  Pythagorean  proposition  we  can  find  the 
chord  of  the  48th,  or  96th,  or  192d  of  a  circumference. 
Multiplying  these  chords  by  48,  or  96,  or  192,  gives  us 
nearly  the  length  of  the  circumference ;  and  the  greater  the 
number  of  times  that  we  bisect  the  arc  of  60°,  the  more 
nearly  will  we  attain  the  exact  length.  The  length  of  the 
semi-circumference,  with  a  radius  of  unity,  or  of  a  circum- 
ference with  a  diameter  of  unity,  is  called  n. 


84  CIRCLES. 

Vastly  more  rapid  ways  of  calculating  n  have  been 
found  by  the  Differential  Calculus.  Its  exact  value  cannot 
be  obtained,  because  the  diameter  and  circumference  are 
not  in  the  ratio  of  any  two  numbers  whatever.  For  ordi- 
nary calculations  it  may  be  taken  as  equal  to  3.1416,  and 
for  the  most  accurate  osculations  3.14159265. 

239.  As  the  area  of  a  circle  is,  by  Art.  237,  equal  to  the  ra- 
dius multiplied  by  the  semi-circumference,  and  as  the  semi- 
circumference  is,  by  Art.  238,  equal  to  the  radius  multiplied 
by  TT,  it  follows  that  the  area  is  equal  to  the  radius  multiplied 
by  itself  and  then  by  n.  In  other  words,  the  area  of  a  circle 
is  3.1416  times  as  large  as  the  square  on  the  radius. 

240.  As  the  square  on  a  radius  is  one  fourth  the  square 
on  the  diameter,  we  may  find  the  area  of  a  circle  by  mul- 
tiplying the  square  of  the  diameter  by  one  fourth  of  3.1416, 
that  is,  by  .7854 ;  or,  for  very  accurate  calculations,  by 
.785398. 

241.  To  find  the  length  of  an  arc  of  any  number  of 
degreeSy  and  of  a  given  radius,  —  The  semi-circumference 
with  a  radius  of  1  is  3.14159265,  and  if  we  divide  this  by 
180  it  gives  the  length  of  one  degree,  .017453.  If  we 
multiply  this  decimal  (.017453)  by  the  number  of  degrees 
in  the  given  arc,  it  will  give  the  length  of  that  arc  in  a 
circumference  whose  radius  is  1.  Multiplying  this  by  the 
given  radius  will  give  the  required  arc. 

242.  To  find  the  area  of  a  sector  of  a  circle  ;  that  is^  of  a 
figure  included  between  two  radii  and  an  arc,  —  Multi- 
ply the  arc  by  the  radius,  and  the  product  will  be  twice 
the  area.  Or  find  the  area  of  the  circle,  divide  it  by  360, 
and  multiply  the  quotient  by  the  number  of  degrees  in 
the  arc. 

243.  To  find  the  area  of  a  segment  of  a  circle  ;  that  is^ 
pf  a  figure  included  between  an  arc  and  its  chord.  —  Find 
the  area  of  the  sector,  having  the  same  arc,  and  also  of  the 
triangle  included  between  the  radii  and  the  chord.    If  the 


DOUBLE   POSITION.  85 

arc  is  less  than  180°,  subtract  the  triangle  from  the  sector; 
if  the  arc  is  more  than  180°,  add  the  triangle  to  the  sector. 

Examples. 

244.  With  a  radius  of  7.3  inches  what  is  the  length  of 
the  circumference  ?  Of  an  arc  of  79°  ?  of  53°  ?  of  58^°  ? 
What  is  the  area  of  the  circle?  Of  a  sector  of  51°?  of 
37°  ?    Of  a  segment  of  63°  ?  of  79°  ?  of  176°  ?  of  183°  ? 

245.  In  measuring  the  altitude  of  triangles,  or  the  dis- 
tance of  any  point  from  a  line,  the  simplest  mode,  justified 
by  the  spirit,  although  not  the  letter,  of  the  usual  postu- 
lates, is  to  place  one  foot  of  the  compasses  on  the  point, 
and  open  them  until  the  other  foot,  swinging  near  the  line, 
will  touch  it  without  crossing  it. 

246.  What  is  the  area  of  a  triangle  whose  sides  are  17, 
23,  31?  What  is  the  area  of  the  four  lots  in  Art.  213? 
Of  the  triangles  in  Arts.  204,  202,  200. 


CHAPTER    VIII. 

DOUBLE    POSITION. 

247.  In  many  practical  problems  there  may  be  no  direct 
method  of  solution,  and  nevertheless  there  may  be  direct 
modes  of  testing  the  accuracy  of  a  solution.  In  these 
cases  the  arithmetical  rule  of  "Double  Position"  is  a  most 
valuable  means  of  obtaining  the  number  sought.  It  pro- 
ceeds upon  the  simple  supposition  that  the  errors  of  a 
result  are  in  proportion  to  the  errors  of  the  data  from 
which  the  result  is  obtained. 

248.  To  solve  a  question  by  double  position,  you  must 
first  discover  a  mode  of  testing  an  answer  by  subjecting  it 
to  calculations  which,  if  the  answer  is  correct,  will  yield  a 


86-  DOUBLE    POSITION-. 

given  number.  Make  two  "  positions,"  or  supposed 
answers,  test  them,  and  note  the  errors  of  the  results. 
Then  the  difference  of  the  results  is  to  the  difference  of 
the  positions  as  the  error  of  either  result  is  to  the  error  of 
its  position,  and  the  solution  of  this  problem  in  the  "  Rule 
of  Three"  will  enable  you  to  correct  your  "position." 

249.  To  solve  the  question  of  double  position  by  geometri- 
cal construction.  —  Draw  a  straight  line  A  B  of  any  length. 
Mark  upon  it,  at  any  convenient 
place,  a  point  C,  to  represent  the 
smaller  number  of  your  two  po- 
sitions. Measure  C  D  equal  to  the  a ^^r t b 

difference  of  your  positions,  taken 
on  the  scale.  Above  or  below  C  and  D,  at  right  angles  to 
A  B,  at  a  distance  equal  to  the  error  of  their  results,  mark 
the  dots  d  and  c  measured  on  the  same  scale  as  C  D,  or  on 
a  different  scale,  and  join  them  by  a  straight  line,  cutting 
A  B  in  E.  Measure  C  E  on  the  same  scale  as  that  on  which 
C  D  was  measured,  and  subtract  it  from  the  smaller  posi- 
tion. This  will  give  a  new  position,  more  correct  than  C 
or  D.  Try  now  two  new  positions,  nearly  equal  to  this 
corrected  one,  and  employ  a  larger  scale  in  constructing 
them,  and  thus  proceed  until  you  find  a  position  nearly 
enough  exact  for  your  purpose.  If  either  result  is  too 
small,  the  corresponding  point  c  or  d  will  be  below  the 
line  A  B,  and  E  will  fall  between  C  and  D,  so  that  C  E 
must  be  added  to  the  position  C.  1^  T>  d  is  less  than  C  c, 
then  E  will  fall  beyond  D,  and  it  will  be  better  to  measure 
D  E,  and  add  it  to  the  position  D.  If  the  line  c  d  is  too 
nearly  parallel  to  A  B,  then  the  distance  CD  must  be 
measured  on  a  smaller  scale,  ox  X>  d  and  C  c  on  a  larger 
scale. 

In  many  cases  a  more  exact  result  can  be  more  rapidly 
attained  by  making  three  positions,  plotting  three  points 
like  d  and  c,  and  then  drawing  an  arc  instead  of  a  straight 


DOUBLE   POSITION.  8t 

line  through  them.  The  intersection  of  this  arc  with  the 
line  A  B  will  then  show  most  exactly  the  true  point  for  a 
"  position  "  which  will  stand  the  test. 

250.  A  few  examples  will  show  more  clearly  the  mean- 
ing of  the  above  directions. 

(a.)  What  two  iiumbers  are  they  whose  sum  is  11  and 
the  sum  of  their  squares  76  ? 

Supposing  the  least  number  to  be  3,  the  greater  will  be 
8,  and  the  sum  of  the  squares  will  be  9  -j-  64  i=  73.  It  is 
therefore  plain  that  3  is  too  large,  or  rather  that  8  is  too 
small.  Supposing,  therefore,  the  least  number  to  be  2.5, 
the  greater  number  will  be  8.5,  and  the  sum  of  the  squares 
will  be  6.25  +  72.25  =  78.5.  On  a  straight  line  A  B,  I 
now  make  two  dots,  C  and  D,  at  the  dis-  Cv 

tance  of  .5  of  an  inch  apart,  because  my  \^    P 

positions  for  the  smallest  number  were  \ 

2.5  and  3,  and  3  —  2.5  =  .5.     Over  C,  I  ^<? 

put  a  dot,  c,  at  the  distance  of  .25  of  an  inch,  because  78.5 
—  76  =  2.5,  and  I  reduce  it  to  one  tenth  the  scale,  for  con- 
venience sake.  Under  D,  I  put  the  dot  c?,  at  the  distance 
.3  of  an  inch,  because  76  —  73  =  3,  which  I  reduce  to  one 
tenth  the  scale.  Joining  c  to  c?  by  a  straight  line,  I  meas- 
ure the  distance  from  C  to  6,  and  find  it  .23  of  an  inch. 
Adding  this  to  2.5  gives  me  2.73  for  a  new  "  position." 

Supposing  successively  the  smaller  number  to  be  2.71, 
2.72,  and  2.73,  the  greater  number  would  become  8.29,  8.28, 
and  8.27  ;  and  the  sums  of  the  squares  76.068,  75.957,  and 
75.846.  The  differences  of  these  numbers  from  the  required 
sum,  76,  are  -f  .063,  —  .043,  and  —  .154. 

Drawing,  now,  the  straight  line  A  B  C,  I  put  the  points 
A  B  C  at  equal  distances,  one  inch  apart ;  that  is,  I  map  the 
differences  between  my  positions,  multiplying  each  by  100. 
I  next  put  the  points  a  b  c  below  and  above  the  line, 
at  the  distances  .315,  .215,  and  .77 ;  that  is,  I  map  the  dif- 
ferences of  my  results,  multiplying  each  by  5.  Holding, 
now,  a   straight  edge  from  A  to  C,  I  mark  the  point  e 


88  DOUBLE   POSITION. 

where  I  judge  by  my  eye  that  an  arc  through  ah  c  would 

cut  the   line  AB. 

Measuring  A  6  I 

find  it  .61,  and  as 

A  B   is  magnified 

100    times,  I   add 

.0061   to  my  first 

position,  giving  me 

2.7161   for  a  new 

position.   Then,  taking  2.7161  and  2.7162  fi)r  new  positions, 

1  might  construct  a  new  one  on  a  scale  of  10,000  for  1, 
instead  of  100  for  1,  and  this  would  give  me  a  result  still 
more  accurate;  that  is,  I  should  find  that  .000018  is  to  be 
added  to  2.7161,  giving  2.716118.  This  process  continued 
would  lead  to  any  desired  degree  of  accuracy. 

(^.)  Find  by  this  process  two  numbers  ichosc  difference 
is  1  and  product  11.  —  Suppose  the  smaller  number  3  and 
larger  4,  the  product  will  be  12,  giving  an  error  of  1.  Sup- 
pose the  smaller  number  2.9  and  larger  3.9,  the  product 
will  be  11.31,  giving  an  error  of  .31.  Map  the  differences 
of  positions  multiplied  by  10,  and  the  difference  of  results 
from  11,  without  changing  the  scale. 

(c.)  Find  two  numbers  whose  difference  is  1  a^id  the  dif- 
ference of  their  squares  is  6.  —  SupjDOse  the  numbers  to  be 

2  and  3,  the  difference  of  their  squares  is  5,  an  error  of  1. 
Suppose  them  to  be  2.2  and  3.2,  the  difference  of  their 
squares  is  5.4,  an  en-or  of  .6.  Suppose  3.5  and  2.5,  the  differ- 
ence of  the  squares  will  be  6.  Hence  3.5  and  2.5  is  the 
exact  answer.  Double  position  frequently  thus  leads,  by  a 
fortunate  guess,  directly  to  an  exact  result. 

{d,)  What  number  is  that^  the  difference  between  whose 
second  and  third  pov^er  sis  12? —  Suppose  2,  and  the  result- 
ing error  is  8.  Suppose  3,  and  the  resulting  error  is  —  6. 
Suppose  2.5,  and  the  resulting  error  is  2.6.  Suppose 
2.6,  and  the  resulting  error  is  1.18.  Suppose  2.7,  and 
the   resulting   error  is  — .39.      Hence  2.5,  2.6,  and  2.7 


INTERPOLATION  AND  AVERAGE.  89 

should  be  mapped  with  their  difference  of  .1  represented 
by  one  inch,  and  the  error  of  the  results  mapped  as  deci- 
mals of  an  inch.  A  curve  drawn  through  the  three  points 
would  give  the  result,  with  an  error  less  than  .001. 

(e.)  What  two  7imnbers  are  they  whose  difference  is  1, 
and  the  difference  of  their  third  powers  7  ? 

(/.)  Solve  the  last  question  when  you  have  written 
8  for  7. 

(</.)  Find  two  numbers  such  that  their  sum  added  t6 
the  square  root  of  their  sum  tcill  equal  12,  and  the  sum 
of  their  cubes  will  equal  189. —  Suppose  3  for  one  number. 
The  cube  of  3  is  27,  which  subtracted  from  189  leaves  162. 
The  cube  root  of  162  is  about  6.6.  The  sum  of  3  and  5.6 
is  equal  to  8.6,  the  square  root  of  which  is  about  2.9. 
Add  2.9  to  8.6  gives  11.5,  and  it  should  give  12,  so  that 
the  error  of  result  is  .5.  A  second  position  will  probably 
give  you  an  exact  result,  so  that  there  will  be  no  need  of 
construction. 

(h.)  Perform  the  same  example^  substituting  176  for 
189. 


CHAPTER    IX. 

INTERPOLATION   AND   AVERAGE. 

251.  Suppose  that  you  wish  to  know  what  is  the  high- 
est point  to  which  the  thermometer  rises  on  a  given  day, 
what  it  is  at  half  past  10  o'clock,  and  what  is  the  average 
heat,  that  is,  the  mean  temperature  of  the  day,  and  at  what 
time  the  heat  is  greatest.  But  suppose  that  you  are  only 
able  to  observe  the  thermometer  at  4  o'clock,  6  o'clock, 
8^  o'clock,  and  so  on,  at  irregular  hours  during  the  day. 
How  shall  you,  from  these  observations,  find  the  answers 
to  your  four  questions  ? 
8* 


90  INTERPOLATION  AND  AVERAGE. 

You  cannot  obtain  perfectly  accurate  answers  in  any 
way ;  but  the  simplest  way  of  getting  tolerably  correct  an- 
swers is  by  geometrical  construction. 

Draw  a  straight  line  A  B,  and  on  it  mark  points  corre- 
sponding in  their  distances  from  each  other  to  the  inter- 
vals of  time  between  your  observations.  The  scale  may 
be  a  quarter  of  an  inch  to  an  hour,  one  tenth  of  an  inch  to 
an  hour,  or  any  other  scale  you  please.  Now,  from  the 
degrees  of  the  thermometer,  at  each  observation,  subtract 
the  greatest  number  of  tens  contained  in  the  lowest  de- 
gree. Set  the  remainders  in  any  convenient  scale,  say  one 
tenth  of  an  inch  to  a  degree,  over  the  corresponding  point 
on  A  B.  Connect  the  points  thus  obtained,  by  drawing 
through  them  as  easy  and  natural  a  curve  as  possible. 
The  highest  jDoint  of  the  thermometer  for  the  day  will  be 
found  by  measuring  the  distance  of  the  highest  point  of 
the  curve  from  the  straight  line  A  B.  The  time  when  the 
thermometer  was  highest  will  be  found  by  measuring  the 
distance  from  A  to  the  point  upon  A  B  under  the  highest 
point  of  the  curve.  The  temperature  at  10^  o'clock  will 
be  found  by  measuring  the  height  of  the  curve  over  the 
point  on  A  B  corresponding  to  10^  o'clock.  The  question 
of  the  mean  temperature,  or  average  height  of  the  ther- 
mometer, will  be  a  little  more  difficult  to  answer. 

Draw,  from  those  points  on  the  line  A  B  between  which 
you  wish  to  find  the  mean  temperature,  perpendicular 
lines,  long  enough  to  pass  through  the  curve,  and  a  little 
more.  Draw  a  fine  silk  thread  tight,  hold  it  on  the  paper 
parallel  to  A  B,  and  move  it  nearer  or  farther  from  A  B 
until  the  area  between  the  curve  and  the  thread,  on  one 
side  the  thread,  seems  equal  to  the  area  between  the 
thread,  the  curve,  and  the  perpendiculars  on  the  other 
side  the  thread  ;  the  distance  of  the  thread  from  the  line 
A  B  will  then  indicate  the  mean  temperature  during  the 
time  included  between  the  perpendiculars. 

252.  It  is  plain  that  the  same  method  can  be  applied 


INTERPOLATION    AND    AVERAGE.  91 

to  any  similar  questions  concerning  the  barometer,  or 
dew-point,  or  other  meteorological  phenomena ;  or  to  the 
force  of  steam,  or  any  thing  varying  by  unknown  laws. 

253.  A  similar  method  may  also  bo  employed  in  calcu- 
lating things  that  alter  by  known,  but  complicated  laws, 
when  you  wish  to  anive  at  an  approximate  result  without 
arithmetical  labor.  Thus  examples  which  we  have  given 
in  the  chapter  on  Double  Position  may  be  solved  by  this 
mode  of  interpolation. 

Let  us  take,  for  example,  the  question.  What  two  num- 
bers differ  by  2,  and  form  the  product  5  ?  This  may  be 
solved  by  double  position,  as  in  Art.  250,  Example  (^),  or  it 
may  be  done  as  directed  in  Art.  251,  by  using  one  of  the 
numbers  as  the  hour,  and  the  product  as  the  temperature. 
If  the  number  subtracted  from  the  products,  to  make  them 
easier  to  plot,  be  taken  equal  to  the  required  product,  5, 
tins  process  of  Art.  251  becomes  exactly  the  same  as  that 
of  Art.  250.  Art.  251  is  therefore  simply  a  wider  and 
more  useful  application  of  the  method  which  you  had  used 
in  double  position. 

254.  I  found  one  morning  the  thermometer  at  6  o'clock 
ten  above  zero,  at  Q^  o'clock  it  was  seven  above,  at  6f 
o'clock  six  above,  at  20  minutes  before  8  it  was  seven 
above,  and  at  8  nine  above.  Now,  when  was  it  coldest, 
what  was  the  greatest  degree  of  cold,  what  was  the  mean 
temperature  for  the  two  hours,  and  what  was  the  temper- 
ature at  sunrise,  namely,  at  20  minutes  past  7  ? 

I  draw  a  line  A  B,  two  inches  long,  to  represent  the  two 
hours,  and  mark  dots 
upon  it  at  distances 
corresponding  to  the 
intervals  between  the 
observations.  Over 
the  ends  of  the  line, 
and  over  the  dots  (finding  the  perpendicular  by  a  square- 


92  INTEEPOLATION  AND  AVERAGE. 

cornered  card),  I  mark  points  as  many  tenths  of  an  inch 
above  AB  as  the  temperature  at  each  observation  was 
above  4°,  six  tenths  above  A,  five  tenths  above  B,  and  so 
on.  Joining  these  points  by  a  curve  which  looks  easy 
and  natural,  I  find  it  approaches  A  B  most  nearly  at  the 
point  D,  1.08  inches  from  A,  and  is  there  .13  of  an  inch 
above  A  B.  Hence  I  know  that  the  greatest  cold  was 
about  five  minutes  past  7,  and  that  the  thermometer  was 
then  at  5.3  above  zero.  At  C,  con-esponding  to  sunrise, 
the  distance  of  the  curve  from  A  B  is  .16  of  an  inch,  and 
therefore  at  sunrise  the  temperature  was  5.6  above  zero. 
The  thread  E  F  appeared  to  leave  a  space  between  itself 
and  the  curve  below  equal  to  that  between  itself,  the  curve 
above,  and  the  perpendiculars  from  A  and  B,  when  it  was 
.29  of  an  inch  above  A  B,  and  therefore  the  mean  tempera- 
ture of  the  two  hours  was  6.9  above  zero. 

255.  The  barometer  stood,  Monday  morning,  at  29.53 
inches,  Monday  evening  29.45,  Tuesday  evening  29.61, 
Wednesday  morning  29.73,  Wednesday  evening  29.71, 
Thursday  evening  29.45,  Friday  morning  29.46.  When 
was  it  highest,  and  when  lowest,  and  what  was  the  aver- 
age height  during  the  four  days  ?  What  was  the  differ- 
ence between  its  lowest  and  highest  points  ? 

In  plotting  this,  the  four  days  might  be  drawn  four 
inches  or  four  half  inches.  The  height  of  the  barometer, 
with  29.40  subtracted,  would  be  in  hundredths  of  an  inch, 
13,  5,  21,  33,  31,  5,  6.  These  may  be  mapped  either  as 
hundredths,  or  fiftieths,  or  even  as  tenths  of  an  inch. 

256.  The  sunlight  at  noon,  Dec.  18,  came  on  my  floor 
2.1  inches  beyond  a  certain  mark ;  Dec.  19,  at  noon,  2.2 ; 
Dec.  20,  at  noon,  2.21 ;  Dec.  21,  at  noon,  2.17 ;  Dec.  22, 
at  noon,  2.07.  At  what  time  was  the  sun  at  his  most 
southerly  position  ? 

257.  Whenever,  in  mapping  observations,  you  map  on 
an  enlarged  scale,  as  when,  in  plotting  the  degrees  of  the 


SURVEYING.  93 

thermometer,  you  represent  them  by  tenths  of  an  incli, 
or,  in  plotting  the  example  of  Art.  256,  you  represent  hun- 
dredths of  an  inch  by  tenths,  you  will  probably  find  that 
no  curve  drawn  through  the  points  will  look  natural  and 
easy ;  or  at  least  none  can  be  drawn  without  some  undu- 
lation. If,  then,  as  in  Art.  256,  we  know  that  the  real 
curve  must  be  without  waves,  we  may  conclude  that  our 
mapping  by  magnifying  the  errors  of  observation  has 
produced  this  appearance.  We  ought,  in  such  a  case,  to 
draw  a  curve  conforming  as  nearly  to  all  the  observations 
as  a  regular  curve  can  —  as  little  above  one  point  or  be 
low  another  as  possible. 


CHAPTER    X. 

SURVEYING. 

258.  Engineers  and  surveyors  require  nice  and  costly 
instruments  for  measuring  angles  and  measuring  lines; 
but  a  boy  who  chooses  to  make  for  himself  a  circle  to 
measure  angles,  and  a  j^ole  to  measure  lengths,  can  read- 
ily do  so  at  a  trifling  cost,  and  find  a  great  deal  of  pleas- 
ure in  doing  it. 

259.  In  measuring  lengths,  the  best  instrument  for  a 
boy  is  a  pole  ten  feet  long,  divided  into  feet,  and  each  foot 
divided  into  tenths.  The  tenths  of  a  foot  will  be  rather 
larger  than  inches,  and  I  recommend  them  in  preference 
to  inches  simply  for  the  ease  of  calculation.  Feet  and 
tenths  of  a  foot  can  be  written  down  like  any  other  deci- 
mal fractions. 

260.  For  measuring  angles,  the  simplest  thin^  for  a 
boy's  use  is  an  instrument  which  he  can  make  for  himself, 
as  follows :  On  a  smooth  piece  of  board,  about  ten  inches 


94  SURVEYING. 

square,  draw  a  circle  with  a  radius  of  five  inches.  Through 
the  centre  of  the  circle  draw  a  diagonal  across  the  board. 
Starting  from  one  of  the  points  where  this  diagonal  crosses 
the  circumference  of  the  circle,  divide  the  circumference 
into  360  equal  degrees.  Make  every  fifth  mark  rather 
longer  than  the  others,  and  number  from  0  around  to  180, 
each  way  from  your  starting  point,  where  the  diagonal 
crosses. 

In  the  diagonal  line  outside  the  circumference,  and  neaf 
the  point  marked  180%  drive  in  a  pin,  which  must  stand 
perpendicular  to  the  surface  of  the  board.  Over  the  other 
end  of  the  line,  near  the  zero,  tack  a  piece  of  card  or  of 
zinc,  with  a  narrow  slit  in  it,  in  such  manner  that  part  of 
the  zinc  or  card  may  stand  up  at  right  angles  to  the  surface 
of  the  board,  and  the  slit  stand  perpendicular  over  the  di- 
agonal. If  now,  looking  with  one  eye  through  this  open- 
ing in  the  zinc,  the  pin  is  made  apparently  to  cover  any 
small  distant  object,  it  is  manifest  that  the  diagonal  line 
will  point  towards  that  object.  Prepare  now  a  flat  stick, 
ten  inches  long,  to  turn  freely  about  a  screw  passed 
through  a  hole  in  its  middle,  and  screwed  into  the  exact 
centre  of  the  circle.  One  end  of  this  stick  must  carry  a 
card,  or  piece  of  zinc,  perforated  with  a  narrow  opening, 
similar  to  that  in  the  stationary  piece.  Next,  bringing 
the  two  pieces  of  zinc,  by  turning  the  stick,  as  near  to  each 
other  as  possible,  look  through  both  slits  at  once,  and  in- 
sert a  pin  in  the  opposite  end  of  the  stick,  so  as  to  hide  the 
pin  already  in  the  board.  A  mark  of  some  kind  (either  a 
pencil  mark  or  the  cutting  of  a  notch)  must  be  made  upon 
the  end  of  the  stick  to  show  the  exact  place  of  the  zero 
division  when  the  stick  is  in  this  position.  ^TsTow,  it  is 
plain  that  if  the  board  is  held  steady  and  immovable,  so 
that  the  stationary  pin,  as  seen  through  the  stationary  slit, 
shall  hide  one  distant  object,  while  the  revolving  pin,  as 
seen  through  the  revolving  slit,  shall  hide  another  distant 


SURVEYING.  95 

object,  the  zero  mark  on  the  end  of  the  stick  will  show 
upon  the  graduated  circle  what  angle  is  made  by  straight 
lines  drawn  from  the  centre  of  the  board  to  the  two  objects. 
This  simple  piece  of  apparatus,  if  made  with  care,  will 
serve  very  well  for  measuring  the  angular  distance  between 
any  two  stationary  objects. 

261.  We  sometimes  wish  to  measure  the  angle  which  a 
line  drawn  from  us  to  an  object  makes  with  a  perpendicu- 
lar or  horizontal  line.  One  mode  of  effecting  this  is  by  an 
artificial  horizon.  The  surface  of  any  fluid  standing  at 
rest  is,  except  near  the  sides  of  the  vessel,  exactly  level. 
Light  is  reflected  from  a  surface  at  the  same  angle  as  that 
at  which  it  falls  upon  it.  For  a  very  high  or  distant  object, 
such  as  the  heavenly  bodies,  we  may  consider  the  light 
that  comes  to  our  eye  as  parallel  to  that  which  falls  upon 
a  saucer  of  ink  and  water  placed  near  the  eye.  For  all 
the  purpose  of  a  boy's  surveying,  we  may  consider  the  top 
of  a  church  steeple,  or  of  a  hill,  as  sufficiently  distant  to 
make  the  same  assumption. 

262.  A  saucer  of  water,  colored  by  any  thing  which  will 
prevent  your  seeing  the  inside  of  the  saucer,  may  be  called 
an  artificial  horizon.  If  you  hold  your  head  in  such 
position  as  to  see  in  the  artificial  horizon  the  image  of  a 
stai'  or  the  vane  of  a  steeple,  and  then,  by  means  of  the 
circle,  measure  the  angle  between  the  star  and  its  image, 
half  that  angle  will  be  the  altitude  of  the  star,  or  the  angle 
which  a  line  drawn  from  the  vane  to  your  eye  makes  with 
a  level  line. 

Thus  let  A  be  a  vane,  B  an  artificial  horizon,  and  C 
the  eye  of  the  beholder.  D  will  be 
the  apparent  image  of  the  vane.  B  E 
is  a  level  line,  and  the  angle  ABE 
is  half  the  angle  A  B  D.  But  if  A  is 
so  distant  from  B  and  C  as  to  make 
the  angle  BAG  insignificantly  small. 


96  SURVEYING. 

A  C  D,  which  is  the  angle  measured  by  the  circle,  may 
be  considered  as  equal  to  A  B  D. 

263.  Instead  of  an  artificial  horizon,  a  plumb  line  may 
be  used ;  but  then  you  will  require  another  instrument,  as 
it  would  be  difficult  to  make  a  plumb  line  hang  from  the 
exact  centre  of  the  circle,  already  occupied  by  the  screw. 

264.  Upon  a  piece  of  board  draw  a  quarter  of  a  circle, 
and  gi-aduate  it  as  for  a  protractor.  At  the  centre  of  the 
arc  drive  in  a  pin  perpendicular  to  the  surface  of  the  board. 
At  the  point  marked  90°  fasten  a  piece  of  card  or  zinc, 
pierced  by  a  narrow  slit.  To  the  pin  tie  a  piece  of  fine 
silk,  long  enough  to  reach  beyond  the  farthest  corner  of 
the  board,  and  to  the  end  of  the  silk  fasten  a  small  weight. 
This  instrument  is  called  a  quadrant.  It  is  plain  that,  if 
the  quadrant  be  held  so  as  to  allow  the  silk  to  play  freely 
over  the  surface  of  the  board,  it  will  mark  upon  the  grad- 
uated arc  the  measure  of  the  angle  which  a  line  drawn 
from  your  eye  (applied  to  the  slit)  to  the  object  hid  by  the 
pin  makes  with  the  silk  thread. 

265.  With  these  four  simple  instruments,  a  ten-foot  pole, 
a  circle,  an  artificial  horizon,  and  a  quadrant,  you  will  be 
able  to  perform  a  great  many  interesting  feats  in  survey- 
ing, and  in  the  measurement  of  heights  and  distances. 

266.  It  is  easier  to  measure  angles  than  to  measure  dis- 
tances, although  it  requires  much  more  care  and  accuracy 
in  order  to  obtain  good  results.  Men  are  therefore  accus- 
tomed to  make  instruments  to  measure  angles  as  perfect  as 
possible,  and  then  in  surveying  to  measure  only  as  many 
lengths  as  is  absolutely  necessary. 

267.  If  you  wish  to  make  a  map  of  a  piece  of  land, 
measure  carefully  the  length  of  one  side,  and  then  all  the 
angles  which  would  be  made  with  this  side  by  diagonals 
drawn  from  its  ends  to  the  different  corners  of  the  field. 
Then  map  it  by  Art.  196.  The  area  of  the  lot  may  be 
found  by  Art.  234. 


HEIGHTS   AND   DISTANCES.  97 

268.  It  may  sometimes  be  more  convenient  to  measure 
the  length  of  more  than  one  side.  I  only  wish  you  to 
remember  that  the  measurement  of  only  one  length  is 
necessary,  and  that  you  will  find  it  easier  to  measure  an 
angle  than  a  line.  You  can,  if  you  please,  after  making  a 
map  by  the  measurement  of  one  side  and  the  angles,  meas- 
ure the  other  sides  on  the  map,  and  then  test  the  accuracy 
of  your  work  by  measuring  them  afterwards  on  the  ground 
with  the  pole. 

269.  In  measuring  a  line,  you  must  remember  that  if 
the  ground  is  uneven  you  will  make  the  line  too  long  if 
you  follow  the  unevennesses  of  the  ground.  You  must 
hold  the  pole  level,  and  mark  the  spot  directly  under  the 
elevated  end.  The  area  of  a  hill-side  is  not  usually  con- 
sidered as  that  of  its  surface,  but  as  that  of  its  base. 


CHAPTER    XI. 

HEIGHTS  AND  DISTANCES. 

270.  To  find  the  height  of  a  i^erpendicular  object  on 
levd  ground^  for  instance^  a  building  or  a  tree,  —  From 
a  point  directly  under  the  top  of  the  object,  measure  off 
any  convenient  distance,  and  then  wdth  the  quadrant  take 
the  altitude^  or  angle  of  elevation.  Then,  by  Art.  195, 
draw  the  triangle,  whose  vertices  are  your  own  eye,  the 
top  of  the  object,  and  a  point  under  the  top  at  the  level 
of  your  eye.  Drawing  beneath  the  base  a  line  to  repre- 
sent the  ground  (the  distance  between  this  line  and  the 
base  being  the  height  of  your  eye),  it  is  manifest  that  you 
can  measure  the  height  of  the  object,  and  also,  if  you 
choose,  the  distance  from  your  feet  to  the  top  of  the 
object. 

9 


98  HEIGHTS   AND   DISTANCES. 

271.  At  a  distance  of  60  feet  from  the  centre  of  a  cliurch 
tower,  I  find  the  angle  of  elevation  of  the  vane  to  be  5G\ 
What  is  the  height  of  the  vane,  my  eye  being  5  feet  above 
the  ground  ? 

At  the  distance  of  15  feet  from  a  wall,  the  angle  of 
elevation  of  its  top  is  63^  If  my  eye  is  5  feet  from  tlio 
ground,  how  long  a  ladder  will  I  need  to  climb  the  wall  ? 

At  the  estimated  distance  of  5  miles,  on  a  level,  from 
the  centre  of  a  mountain,  the  image  of  its  summit  in  an 
artificial  horizon  is  13°  below  the  real  summit.  Wliat  is 
the  j^robable  height  of  the  mountain  above  the  point  where 
I  stand  ? 

272.  From  a  hnovm  height  to  find  the  distance  of  an 
object  on  level  ground,  —  If  we  look  in  the  opposite  direc- 
tion on  the  quadrant,  that  is,  apply  our  eye  near  the  centre, 
and  make  the  pin  appear  in  the  centre  of  the  slit  at  the 
moment  when  we  can  see  the  'given  object  through  the 
slit,  it  is  manifest  that  the  thread  wdll  mark  the  angle  of 
elevation  which  w^e  should  have  if  viewed  from  the  place 
of  the  object.  Therefore,  in  this  problem,  the  constniciion 
is  the  same  as  that  of  Art.  269,  except  that  the  lieiglit  is 
given,  and  the  angle  to  be  used  is  found  by  subtracting  the 
observed  angle  from  90°. 

From  the  top  of  Prospect  Hill,  which  I  know  to  bo  420 
feet  above  the  level  of  a  plain  below,  I  observe,  with  the 
quadrant,  a  stone  wall  on  that  plain  to  be  7°  below  a  level. 
What  is  the  distance  of  the  wall  from  the  top  of  the  hill 
in  a  straight  line,  and  what  is  its  distance  on  a  level  ? 

Another  wall  is  seen  directly  between  me  and  the  first 
wall,  at  a  depression  of  13°.  What  is  the  distance  between 
these  two  walls  ? 

273.  To  solve  Art,  270  when  a  point  under  the  top  of  the 
object  is  inaccessible,  —  This  is  frequently  the  case,  espe- 
cially with  mountains  and  hills.  The  simplest  mode  of 
solution  is,  to  measure  on  level  ground  a  straight  line  going;i. 


HEIGHTS   AND   DISTANCES. 


99 


directly  towards  the  hill,  and  take  the  angle  of  elevation 
at  each  end  of  this  line ;  then  drawing  a  straight  line,  A  D, 
measure  off  at  one  end  a 
portion  A  B  to  represent 
your  measured  line.  Draw- 
ing next  the  two  lines,  A  C 
and  B  C,  at  the  proper 
angles,  to  represent  rays  of  light  coming  from  the  object, 
their  intersection,  C  will  represent  the  place  of  the  object, 
and  the  perpendicular  distance  from  the  line  A  D  can 
be  easily  measured. 

What  is  the  perpendicular  height  of  a  hill  when  the 
measured  level  line  is  310  feet,  and  the  angles  27°  and  15°  ? 

274.  If  level  ground  cannot  be  obtained  for  the  meas- 
urement, of  Arts.  270  and  271,  it  is  only  necessary  to 
measure  by  the  quadrant  the  line  of  elevation  of  the 
ground  on  which  the  measurement  is  made.  This  can  be 
done  by  placing  at  the  highest  end  of  the  measured  line  a 
stake,  whose  toj)  shall  be  on  a  level  with  your  eye  when 
you  stand  at  its  side.  The  angle  of  elevation  of  the  top 
of  this  stake,  taken  from  the  other  end  of  the  line,  will  be 
the  elevation  of  the  ground. 

275.  The  case  supposed  in  Art.  270,  and  that  in  Art. 
273,  are  in  geometry  the  same,  only  that  in  Art.  270  one  of 
the  angles  of  elevation  is  90°. 

276.  In  constructing  a  case  under  Art.  274,  draw  a  line 
A  B  to  represent  a  level  drawn  through  the  station  most 
distan t  fro m  th  e  obj  e ct.  Next 
draw  a  line  A  C  peiT^endicu- 
lar  to  A  B,  to  represent  the 
height  of  your  eye.  Draw 
now  A  D  and  C  E  parallel  to 
each  other,  A  D  representing 
the  ground,  and  the  angle 
BAD  being  the  angle  of 
elevation    of    the     ground. 


I 


100  MISCELLANEOUS   EXAMPLES. 

Draw  a  line  through  C,  making  an  angle  with  A  B  equal 
to  the  elevation  of  the  object  from  C.  Through  H,  the 
top  of  the  stake,  draw  a  line  at  the  proper  elevation,  and 
the  intersection  F  with  the  former  line  will  represent  the 
object.  From  this  point  let  fall  a  perpendicular  on  A  B, 
and  you  may  measure  on  that  perpendicular  the  height  of 
the  object,  above  or  below,  either  the  ground,  or  the  level 
AB. 

277.  Measuring  down  a  steep  hill-side  a  line  220  feet 
long,  directly  towards  a  church  steeple,  which  I  knew  to 
be  123  feet  high,  I  found  the  angle  of  depression  of  the  vane, 
at  the  upper  end  of  the  line,  to  be  10°,  and  at  the  lower  end 
4°.  The  depression  of  the  line  on  the  hill-side  was  23°. 
How  high  was  the  upper  end  of  the  line  above  the  foot  of 
the  steeple  ? 

278.  At  the  foot  of  a  hill,  which  rises  at  an  angle  of  17°, 
the  top  of  a  tree  on  the  hill-side  has  an  angle  of  elevation 
equal  to  37°.  On  measuring  my  distance  to  the  foot  of 
the  tree,  I  find  it  about  417  feet.  What  is  the  height  of 
the  tree  ? 

279.  A  house  or  tower  may  be  taken  as  the  side  of  the 
hill  on  which  the  base  line  is  measured,  in  which  case  the 
angle  of  its  elevation  is  90°. 


CHAPTER    XII. 

MISCELLANEOUS   EXAMPLES. 

280.  Prove  that  the  two  diagonals  of  a  rectangle  are 
equal,  and  that  they  mutually  bisect  each  other.* 

281.  Prove  that  perpendiculars  let  fall  from  any  two 
points  in  a  straight  line  upon  a  parallel  straight  line  are 
equal. 

282.  Given  a  side  of  a  triangle  21  feet,  the  opposite 


MISCELLANEOUS    EX;Ai!I*Lt:s.  101 


angle  20°,  and  the  ratio  of  the  adjacent;  •Jlligtes  3  :1,  wlint' 
are  the  other  sides  ?  and  the  area  ? 

283.  One  side  of  a  triangle  is  40  feet,  the  opposite  angle 
is  obtuse  (greater  than  90°),  and  the  three  angles  are  in 
the  proportion  of  31,  20,  and  9.  Required  the  other  sides, 
and  the  area. 

284.  Prove  that  a  square,  circumscribed  about  a  circle, 
is  twice  as  large  as  an  inscribed  square. 

285.  Prove  that  the  angle  between  two  chords  that 
do  not  touch  each  other  is  measured  by  half  the  difference 
of  the  arcs  between  them.  (Prolong  them  until  they 
meet,  join  their  alternate  extremities  by  a  third  chord,  and 
aj^ply  Arts.  144  and  145.)     Prove  it  also  in  another  mode. 

286.  Prove  that  the  angle  made  by  a  chord  and  a  tan- 
gent is  measured  by  half  the  difference  of  the  arcs  between 
them.  Show  what  this  becomes  when  the  point  of  tan- 
gency  is  at  one  end  of  the  chord,  and  when  the  chord 
also  becomes  a  diameter. 

287.  When  two  chords  cross,  what  is  the  measure  of 
their  angle  ? 

288.  Parallel  lines  intercept  equal  arcs.  Prove  it  by  Art. 
285. 

289.  Prove  that  a  perpendicular,  let  fall  from  the  centre 
of  a  circle,  on  a  chord,  bisects  it. 

290.  Two  circles  intersect.  Given  their  radii,  15  and 
10,  and  the  length  of  their  common  chord,  7,  what  is  the 
distance  of  their  centres  ? 

291.  Prove  that  a  quadrangle,  whose  noith-east  angle 
measures  70°,  and  south-west  angle  measures  110^,  can  be 
inscribed  in  a  circle.  (Divide  it  by  a  north-west  and  south- 
east diagonal,  and  prove  that  the  circle  which  is  circum- 
scribed about  one  triangle  is  circumscribed  about  the  otlier.) 

292.  Substitute  85°  and  95°  in  Art.  291;  add  that  a  third 
angle  is  100°,  and  determine,  by  construction,  the  ratio  of 
each  side  to  the  diameter  of  the  circle, 

9* 


102  MISCELT-ANEOUS    EXAMPLES. 

293.  Two  ciiords'  intersect.  The  segments  nearer  the 
centre  of  the  ch*cle  are  equal ;  j^rove  that  the  chords  inter- 
cept equal  arcs. 

294.  Prove  that  parallel  lines  intercepted  between  par- 
allel lines  are  equal. 

295.  How  is  the  angle  between  two  tangents  to  be 
measured  ? 

296.  Two  sides  of  a  triangle  being  given,  and  the  length 
of  a  perpendicular  let  fall  on  one  from  the  opposite  vertex, 
to  construct  the  triangle  ? 

297.  Two  sides  being  given,  41  and  53,  and  the  altitude 
of  the  triangle  from  the  side  53  being  30°,  what  is  the  third 
side? 

298.  My  house  lot  is  a  triangle,  with  a  front  of  100  feet. 
The  perpendicular  distance  to  the  back  corner  is  67  feet. 
The  perpendicular  distance  from  one  end  of  the  front  line 
to  the  opposite  side  is  79  feet.    Draw  a  plot. 

299.  Given  the  base  of  a  triangle,  and  its  altitude,  also 
the  altitude  when  another  side  is  taken  as  base.  Construct 
the  triangle. 

300.  One  side  of  a  triangle,  and  the  perpendiculars  let 
fall  from  its  extremities  on  the  other  sides,  being  given,  to 
construct  the  triangle. 

301.  One  angle  of  a  triangle  is  70^,  and  the  perpen- 
diculars let  drop  from  the  other  vertices  on  the  sides 
are  40  and  30  feet.     Construct  the  triangle. 

302.  Announce  problem  300  in  general  form,  and  give 
a  written  exact  rule  for  its  solution. 

303.  One  angle  of  a  triangle  is  given,  and  the  length  of 
a  perpendicular  let  fall  from  its  vertex  on  the  opposite 
side ;  also  the  perpendicular  let  fall  from  another  vertex. 
Construct  the  triangle. 

304.  Given  the  base  of  a  triangle,  an  adjacent  angle, 
^nd  its  altitude.     Construct  the  triangle.  \ 

305.  On  a  given  line,  as  a  chord,  construct  au  arc  of  a 


MISCELLANEOUS    EXAMPLES.  103 

i^ivon  number  of  degrees.     Use  Arts.  289  and  286  to  find 
the  centre  of  the  arc. 

80G.  Given  the  base  of  a  triangle,  its  opposite  angle,  and 
the  altitude.     Construct  the  triangle,  using  Art.  305. 

307.  On  the  three  sides  of  any  triangle  describe  squares 
exterior  to  the  triangle.  Connect  the  outermost  corners 
of  the  squares  by  right  lines,  and  prove  that  each  of  the 
three  triangles  thus  formed  is  equivalent  in  area  to  the 
original  triangle. 

308.  A  quadrilateral,  with  diagonals  equal  to  12  and  14, 
is  inscribed  in  a  circle  whose  radius  is  8.  The  diagonals 
make  an  angle  of  78°  with  each  other.  Construct  the 
quadiilatcral.  Give  a  general  rule.  Show  when  the 
problem  would  be  impossible. 

309.  Given  one  angle  of  a  triangle,  and  the  segments  of 
the  opposite  side  made  by  a  perpendicular  let  fall  from  the 
vertex.     Construct  the  triangle. 

310.  A  dressed  piece  of  timber  is  8  inches  by  6,  and  is 
3  feet  long.  What  is  the  diagonal  on  the  end,  on  the  side, 
and  on  the  edge  ?  What  is  the  longest  straight  line  that 
could  be  passed  through  the  timber  ? 

311.  A  solid  of  six  rectangular  faces,  like  that  of  Art.  810, 
is  a  rectangular  parallelopiped.  Given  its  three  dimensions, 
find  by  construction  its  diagonal. 

312.  Find  by  construction  the  value  of  such  surds  as 
the  square  root  of  19,  or  of  29,  or  39,  or  79,  or  17,  or  24,  &c, 

313.  The  base  of  a  triangle  is  20  feet,  the  opposite  angle 
30°,  and  the  distance  from  the  middle  of  the  base  to  the 
opposite  vertex  is  18  feet.     Construct  the  triangle. 

314.  Announce  Art.  313  in  general  form,  give  a  general 
solution,  and  state  the  cases  of  impossibility. 

315.  One  angle  of  a  triangle,  adjacent  to  the  base,  is 
SO"* ;  the  altitude  is  20  feet ;  the  diameter  of  the  circum- 
scribed circle  is  40  feet.     Construct  the  triangle. 

316.  With  the  same  elements  given  as  in  Art.  315, 
when  would  the  problem  be  impossible  ? 


104  MISCELLANEOUS   EXAMPLES. 

317.  When  two  circles  are  tangent  to  each  other,  prove 
that  the  point  of  tangency  is  in  a  straight  line,  joining 
the  centres. 

318.  Prove  that  a  perpendicular  from  any  point  in  a  cir- 
cumference, let  fall  on  a  diameter,  is  a  mean  proportional 
between  the  segments  of  the  diameter. 

319.  Find  by  construction  a  mean  proportional  between 
two  given  lines. 

320.  A  chord  prolonged  until  it  reaches  a  point  without 
a  circle  may  be  called  a  secant.  Prove  that  if  two  secants 
are  drawn  to  one  point,  the  entire  secants  are  in  the  inverse 
ratio  of  the  parts  outside  the  circle. 

321.  If  a  tangent  and  secant  are  drawn  to  one  point, 
prove  that  the  tangent  is  a  mean  proportional  between  the 
whole  secant  and  the  part  outside  the  circle. 

322.  Prove  that  three  points  fix  the  position  of  a  plane. 
(Imagine  the  plane  rotating  on  two  of  the  points.  Can  it 
have  more  than  one  position,  and  still  include  the  third 
point  ?) 

323.  The  intersection  of  two  planes  is  a  straight  line. 
Prove  by  Art.  12. 

324.  A  sphere  is  a  solid  whose  surface  is  at  every  point 
equidistant  from  a  point  called  the  centre.  Prove  that  a 
section  by  a  plane  passing  through  the  centre  is  a  circle. 
Also  prove  that  a  section  of  a  sphere  by  any  plane  is  a 
circle. 

325.  The  railroad  starts  due  north  from  Tipton,  and 
runs  straight  one  mile,  then  curves  to  the  east  with  a 
radius  of  1320  feet  until  it  bears  37°  east  of  north,  and 
runs  straight  in  that  direction  1^  miles  ;  then  curves  west- 
erly, with  a  radius  of  660  feet,  until  it  bears  17°  west  of 
north,  and  runs  straight  in  that  direction  3 J-  miles ;  then 
curves  easterly  on  a  radius  of  2640  feet,  until  it  runs  3^ 
east  of  north,  and  runs  straight  2  miles,  to  Haworth. 
Draw  a  map,  determine  the  length  of  the  railroad,  and  tell 


MISCELLANEOUS   EXAMPLES.  105 

the  distance  and  direction,  in  an  air  line,  of  Ilaworth  from 
Tipton. 

326.  The  first  straight  line  starting  from  Tipton  rises  40 
feet ;  the  first  curve  rises  at  the  rate  of  30  feet  to  the  mile ; 
the  second  straight  line  runs  level  for  five  eighths  of  a  mile, 
and  rises  the  rest  of  the  way  at  the  rate  of  25  feet  to  the 
mile;  the  second  curve  is  level;  the  third  straight  lino 
rises  the  first  mile  45  feet,  descends  for  three  fourths  of  a 
mile  at  the  rate  of  48  feet  a  mile,  and  rises  the  rest  of  the 
way  at  the  rate  of  30  feet  to  the  mile ;  the  third  curve 
rises  for  two  thirds  its  length,  at  the  rate  of  35  feet  a  mile, 
and  the  other  third  descends  at  the  rate  of  40  feet  a  mile ; 
the  fourth  straight  line  is  descending  all  the  way,  at  30  feet 
to  the  mile.  Draw  a  profile  on  a  horizontal  scale  of  1  mile 
to  an  inch,  and  vertical  scale  of  100  feet  to  an  inch,  and 
tell  the  difierence  of  level  between  Tipton  and  Haworth. 


106  SOLID   GEOMETRY, 


PART  III. 

SOLID    GEOMETRY. 


PREFATORY   NOTE. 

In  this  Third  Part  we  have  removed  the  figures  from 
the  text,  in  order  to  give  the  student  the  opportunity  to  use 
his  geometric  imagination.  By  steadily  fixing  the  points 
named  in  his  imagination,  he  may  frequently  dispense  with 
the  figure.  But  in  recitation  he  should  be  able  to  draw 
his  own  figure.  And  in  private  study  let  him  draw  his 
own  figure  if  he  does  not  clearly  conceive  it  without 
drawing.  In  unsolved  problems  he  will  frequently  be 
obliged  to  use  his  pencil,  in  order  to  gain  a  clearer  concep- 
tion of  the  problem.  If  he  can  neither  dispense  with  a 
figure,  nor  draw  one  himself,  let  him  turn  to  the  plate  at 
the  end  of  the  volume. 

In  some  cases  the  notation  of  the  points  saves  the  neces- 
sity of  a  figure.  Thus,  in  reasoning,  in  Art.  412,  concerning 
two  solid  bodies,  each  with  four  solid  corners,  we  have 
denoted  the  corners  of  one  by  a,  Z>,  b\  c',  and  of  the  other 
by  a\  Z>,  5',  c' ;  and  thus  the  notation  shows  that  three  cor- 
ners in  one  coincide  with  three  in  the  other,  and  that  the 
fourth,  a,  is  analogous  in  position  to  the  fourth,  a'^  in  the 
other. 


RATIO    AND    rROPOKTION.  107 

CHAPTER   I. 

RATIO    AND    rilOrOUTION. 

327.  In  algebraic  notation  letters  are  used  to  represent 
numbers,  cither  known  or  unknown,  and  the  results  of 
arithmetical  operations  on  those  numbers  are  represented 
by  signs. 

328.  The  sum  of  the  two  numbers  a  and  x  is  written 
a  -j-  i^*5  and  is  called  a  plus  x, 

329.  The  difference  between  a  and  x  is  written  a  —  a*, 
and  is  called  a  minus  x, 

330.  The  product  of  a  by  x  is  written  cither  a  X  ^t  or 
a  •  a*,  or  simply  a  x^  and  is  called  simply  a^  x, 

331.  The  quotient  of  a  divided  by  x  is  written  a  -^  cc, 

a 
or  a  :  x.  or  — . 

X 

332.  A  power  or  root  is  written  either  by  means  of  ex- 
ponents or  of  the  radical  sign.     Thus,  the  ccth  power  of  a  is 

1  rr . 

cC^  and  the  x\h  root  of  a  is  either  a^  or  V  a, 

333.  A  bar  over  two  quantities  indicates  that  they  are 
to  be  considered  together,  and  a  parenthesis  is  used  for  the 
same  purpose.  Thus,  Va  +  x  is  the  sum  of  x  and  of  the 
square  root  of  a ;  but  ^  a-\-x  is  the  square  root  of  the 
sum  of  a  and  x. 

334.  The  notation  thus  far  explained  may  be  illustrated 
by  an  example;  such  as  '^ i{a xr- -\- {a yY  —  tf -r- tS- 
Here  the  number  a  must  be  multiplied  by.  the  second 
power  of  ic,  and  the  product  added  to  the  second  power 
of  the  product  a  times  y.  From  this  sum  we  must  sub- 
tract the  number  U     The  cube  root  of  this  difference  must 


108  tlATIO   AND   PROPOETION. 

be  raised  to  the  second  power,  and  then  divided  by  the 
number  %  and  finally  the  square  root  of  the  quotient  must 
be  extracted. 

335.  The  sign  z=l  signifies  that  the  sums  of  the  quan- 
tities on  either  side  of  it  are  numerically  equal.  Thus, 
V zzmt-^m  signifies  that  the  number  P  is  equal  to  the 
sum  of  m  and  the  product  n  times  L 

336.  The  signs  >  and  <^  are  signs  of  inequality.  Thus, 
P  >  Q  and  Q  <^  P  signify  that  P  is  greater  than  Q,  and 
Q  is  less  than  P. 

337.  The  first  letters  of  the  alphabet  usually  signify 
known,  and  the  last  letters  unknown,  quantities. 

338.  The  signs  =,  >,  <^,  are  the  only  verbs  in  alge- 
braic language,  so  that  each  sentence  must  contain  one  of 
them.  Such  a  sentence  is  called  an  equation.  Equations 
containing  the  sign  >,  or  <;,  are  sometimes  called  ine- 
qualities. 

339.  An  equation  may  be  transposed  in  any  form  what- 
ever, if  we  are  but  careful  to  preserve  the  equality  of  the 
two  members;  that  is,  to  add  or  subtract  from  one  side  of 
the  sign  precisely  what  we  add  or  subtract  from  the  other, 
&c.     Thus,  suppose  we  have  the  equation  v 

and  wish  to  find  the  value  of  x  in  terms  of  t  and  a.  We 
may  first  extract  the  square  root  of  each  member,  which 
simply  gives  a-  —  ta=:x^  —  t. 

We  may  now  add  t  to  each  member,  producing  the  equa- 
tion a^  —  j5a  +  ^  =  cc^. 
From  this  we  may  at  once  infer  that 

x'^=za'^  —  ta-^-t ; 
and  extracting  the  square  of  each  member,  obtain 

340.  The  quantity  under  the  radical  sign  may  be  put 
jnto  other  forms,  thus :  — 


RATIO   AND    PROPORTION.  109 


x=:  \/  a^-\-t  —  at. 


aj=^a^+(l  —  a)t. 
x=i  yj  a^  —  {a  —  1)^. 
an:  ^  a{a  —  0+^- 


x=^J  t  —  (t  —  a)  a. 
341.  The  equality  of  two  ratios  is  called  a  proportion. 
Thus,  the  proportion  a  is  to  c  as  A  is  to  C  is  announced 
also  by  saying  that  the  r;jtio  of  a  to  c  is  the  same  as  that 
of  A  to  C.  Writing  this  as  an  equality  between  two  quo- 
tients, we  obtain,  — 
a      A 

(^•)  c  =  C 

Multiplying  both  members  by  the  quantity  C  c,  we  get,  — 

(2.)  aC=iAc, 

Dividing  by  A  C  will  then  give  us 
a       c 

(^•)  A  =  C 

Adding  A  a  to  both  members  of  (2.),  we  obtain, — 

(4.)  Aa  +  Caiz:Aa  +  Ac. 

Multiplying  each  member  of  (2.)  by  2,  and  subtracting  the 
product,  member  by  member,  from  (4.),  gives,— 

(5.)  A  a  —  C  a  =  A  a  —  A  c. 

Equations  (4.)  and  (5.)  may  be  divided  into  factors,  and 
written  as  in  Art.  340. 

(6.)  «(A+C)  =  A(«  +  c). 

(7.)  a(A  — C)=:A(a  — c). 

Dividing  (6.)  by  (  A  +  C  )  (  a  +  c  ),  and  (7.)  by  (A  —  C  ) 
(a  —  c ),  will  give  us,  — 

a     A 

.     (^•)             "^+^~A  +  C 
a A 

(^•^        "^ir^~"A— c 

Hence,  by  equation  (3.),  we  obtain,  — 
10 


a 

a  +  c 

c 

A" 

a 

"A+C" 
a —  c 

"C 
c 

110  PLANES    AND    ANGLES. 


(10.) 

("•)  A-A-C-C 

Adding  a  c  to  both  members  of  (2.),  gives  us,  --- 

(12.)  ac-|-aC=iac-f-Ac. 

Which,  divided  into  factors,  becomes 

(13.)  a(c  +  C)z:rc(a  +  A). 

And  this  divided  by   c(c-f-C)  furnishes   us  with  the 
equation,  — 

a      a-X-  K 

A  great  variety  of  results  may  thus  be  obtained  from 
the  primitive  equation  (1.),  all  of  which  may  evidently  bo 
of  use  in  treating  the  subject  of  similar  figures  in  Geometry. 


CHAPTER    II. 

PLANES    AND   ANGLES. 

342.  If  of  two  straight  lines,  having  a  point  in  common, 
and  at  right  angles  to  each  other,  the  first  remain  station- 
ary as  an  axis,  while  the  second  revolves,  the  second  will 
generate  a  surface  called  a  plane,  and  the  first  will  be  a 
line  at  right  angles  to  the  plane ;  that  is,  a  perpendicular 
to  the  plane. 

343.  It  is  manifest  that  all  points  in  the  plane  will  be 
equidistant  from  any  two  points  in  the  perpendicular  taken 
equidistant  from  its  foot.  A  plane  may  be  defined  from 
the  converse ;  that  is,  a  surface,  every  point  of  which  is 
equidistant  from  two  given  points,  is  a  plane. 

344.  Any  two   points   in   a  plane   being  joined   by  a 


PLANES   AND   ANGLES.  Ill 

straight  line,  every  part  of  that  line  is  in  tlie  plane. 
Proof,  Let  the  foot  of  a  perpendicular  be  1*,  the  given 
points  be  B  and  C,  and  let  A  and  A'  be  points  in  the  per- 
pendicular equidistant  from  P.  The  triangles  ABC  and 
A'  B  C  are  equal,  because  their  sides  are  equal.  Let  D  be 
any  point  in  the  line  B  C.  Then  the  triangles  A'  B  D  and 
A  B  D  have  the  sides  A  B  =  A'  B,  and  B  D  common,  and 
the  angles  at  B  equal ;  whence  A  D  =  A'  D,  and  the  point 
D  is  in  the  plane. 

345.  Three  points  fix  the  position  of  a  plane.  For  if  a 
plane,  passing  through  two  of  the  points,  be  swung  round 
upon  the  line  joining  them,  as  an  axis,  it  can  evidently 
take  but  one  position,  including  the  third  point. 

346.  Two  parallel  lines  are  of  necessity  in  one  plane. 
For  if  through  any  point  in  the  second  line  we  draw  a 
line  parallel  to  the  first  line,  and  in  the  same  plane  with  it, 
it  must  coincide  with  the  second  line ;  therefore,  the  sec- 
ond line  is  in  that  plane. 

347.  If  a  straight  line  move  in  such  manner  that  any 
two  points  in  it  move  in  parallel  straight  lines,  it  generates 
a  plane. 

348.  Two  lines,  having  a  point  in  common,  lie  in  one 
plane.  For  a  straight  line  from  any  point  in  the  second 
line  drawn  through  its  intersection  with  the  first  line,  and 
in  the  same  plane  with  the  first  line,  must  coincide  with 
the  second  line. 

349.  A  perpendicular  to  two  lines  at  their  point  of  inter- 
section is  perpendicular  to  their  plane ;  that  is,  by  Art.  1, 
is  perpendicular  to  every  line  in  the  plane  drawn  through 
its  foot.  Proof.  Using  the  notation  of  Art.  344,  let  B  and 
C  be  chosen  in  the  given  lines,  and  we  have  only  to  show 
that  the  angle  D  P  A  is  a  riglit  angle.  But  this  follows  at 
once  from  the  fact  that  in  the  triangle  A  A'  D  we  have 
A  D  1=  A^  D,  and  the  base  A  A'  is  bisected  at  P. 

350.  When  a  line  is  perpendicular  to  a  plane,  the  plane 
is  also  said  to  be  perpendicular  to  the  line. 


112  PLANES    AND   ANGLES. 

851.  It  is  manifest  from  the  Pythagorean  proposition, 
that  a  perpendicular  measures  the  shortest  distance  from 
a  point  to  a  plane,  and  that  of  two  lines  from  a  point  to  a 
plane,  that  more  nearly  perpendicular  is  shorter. 

352.  The  intersection  of  two  planes  is  a  straight  line. 
For,  by  Art.  344,  the  straight  line  joining  two  points  of 
the  intersection,  lies  wholly  in  both  planes. 

353.  A  second  plane  including  a  perpendicular  to  the 
first  is  said  to  be  perpendicular  to  the  first. 

354.  If  a  straight  line  be  drawn  in  the  first  plane,  from 
the  foot  of  the  perpendicular,  at  right  angles  to  the  inter- 
section, it  will  be  at  right  angles  to  two  lines  in  the  second 
plane,  and  be  a  perpendicular  to  it.  Hence,  each  plane  is 
perpendicular  to  the  other,  and  they  are  said  to  be  at 
right  angles  to  each  other. 

355.  The  angle  made  by  two  planes  may  be  called  a 
diedral  angle.  A  diedral  angle  is  measured  by  the  angle 
made  by  two  lines,  one  in  each  plane,  each  perpendicular 
to  the  intersection  of  the  planes.  For  it  is  manifest,  that 
if  the  planes  be  brought  to  coincidence,  these  lines  coin- 
cide, and  that  if  the  planes  be  then  swung  open  the  angle 
of  these  lines  is  generated  with  exactly  the  velocity  of  the 
motion  of  the  planes. 

356.  Parallel  lines,  making  the  angle  zero,  may  be  con- 
ceived as  meeting  at  an  infinite  distance  in  either  direction. 
In  like  manner,  when  a  diedral  angle  is  zero,  the  planes 
do  not  of  necessity  coincide,  but  are  parallel,  having  their 
intersection  at  an  infinite  distance. 

357.  As  the  intersection  of  parallel  planes  may  be  at 
an  infinite  distance  in  any  direction,  any  two  parallel  lines, 
of  which  one  is  in  either  plane,  may  be  considered  as 
measuring  their  angle. 

858.  A  line  parallel  to  a  line  in  a  plane  is  said  to  be 
parallel  to  the  plane. 

859.  A  straight  line,  neither  perpendicular  to  a  plane 


PLANES    AND    ANGLES.  113 

nor  parallel  to  it,  makes  an  angle  Avitli  it,  said  lo  bo 
equal  to  the  angle  which  the  line  makes  with  the  intersec- 
tion of  the  plane  by  a  perpendicular  plane,  including  the 
given  line. 

360.  When  two  parallel  planes  are  cut  by  a  third,  the 
intersections  are  two  parallel  lines.  They  are  straight  lines 
by  Art.  352,  and  although  in  the.  same  plane,  cannot  ap- 
proach each  other  in  either  direction,  because  the  inter- 
section of  the  parallel  planes  is  at  an  infinite  distance,  in 
any  direction. 

361.  A  straight  line  makes  the  same  angle  with  either 
of  two  parallel  planes,  whether  the  angle  be  zero,  a  right 
angle,  or  of  intermediate  value. 

362.  Parallel  lines  intercepted  between  parallel  planes 
are  equal.  For,  joining  the  points  of  interception  by 
straight  lines  in  the  planes,  gives,  by  Art.  360,  a  parallelo- 
gram, whose  sides  are  of  course  equal. 

Hence  it  is  evident  that  parallel  planes  are  every  where 
equidistant. 

363.  When  two  lines  neither  intersect  nor  are  parallel, 
it  may  be  made  evident  by  Art.  351  that  the  shortest  dis- 
tance between  them  is  the  distance  of  a  point  in  one  from 
a  plane  parallel  to  it  drawn  through  the  other.  Hence, 
when  two  lines  neither  intersect  nor  are  parallel,  it  is  evi- 
dent that  the  right  line  joining  their  points  of  nearest 
approach  is- perpendicular  to  each. 

364.  If  two  right  lines  are  intercepted  between  parallel 
planes,  a  third  parallel  plane  will  divide  the  intercepts  in 
the  same  proportion.  Proof,  Let  A,  C,  and  A',  C^,  be  tho 
points  of  intersection  of  the  right  lines  with  the  first  planes, 
and  B,  B',  the  points  of  intersection  with  the  third  plane. 
Draw  A  B''  C^'  parallel  to  A'  C,  and  Ave  have,  by  Art.  362, 
A  B"  =  A'  B'  and  B''  0"  =  B'  C^  Completing  the  simi- 
lar triangles  ABB"  and  A  C  O'  gives  us  A  B  :  B  C  = 
AB''  ;B'  C    :=  A'B'  :B'C'. 

10* 


114  PLANES    AND    ANGLES. 

365.  The  intersection  of  three  planes  produces  triedral 
angles,  the  point  common  to  the  three  planes  being  called 
their  vertex.  In  English,  the  vertex  of  a  diedral  angle  is 
usually  called  an  edge,  that  of  a  triedral  angle  a  solid 
corner. 

366.  If  a  third  plane  is  j^erpendicular  to  each  of  two 
planes,  it  is  2:)erpendicular  to  their  intersection.  Proof, 
From  the  triedral  vertex  raise  a  perpendicular  to  the  third 
plane,  and  as  it  must  be  in  both  the  other  planes,  it  will 
coincide  with  their  line  of  intersection. 

367.  The  vertex  of  a  triedral  angle  is  the  vertex  of 
three  plane  angles,  constituting  the  faces  {edrai)  of  the 
triedral  angle. 

368.  The  sum  of  any  two  of  the  angles  of  the  faces  is 
greater  than  the  third.  For,  if  they  were  simply  equal, 
the  two  faces  would  be  brought  into  the  same  plane  with 
tlie  third,  and  thus  reduce  the  solid  corner  to  one  plane ; 
and  if  the  sum  of  two  were  less  than  the  third,  the  solid 
corner  would  be  impossible. 

369.  If  two  triedral  angles  have  the  same  angles  on 
the  fices,  the  diedral  angles  between  equal  plane  angles 
are  equal.  Proof.  Let  A  be  the  vertex  in  one,  A^  in  the 
other,  and  the  plane  angles  B  A  C,  B  A  D,  D  A  C,  be  re- 
spectively equal  to  B'  A'  C,  B'  A'  D',  and  D'  A'  C.  Make 
A'  B'  ~  A  B,  and  the  angles  A  B  C,  A  B  D,  A'  B^  C^ 
A'  B'  D'  all  right  angles.  We  have  now  to  prove  that  the 
angle  D  B  C  equals  the  angle  D'  B'  C,  —  which  is  done  if 
we  prove  that  the  triangle  D  B  C  equals  the  triangle 
J}'  B'  O.  But  the  triangles  ABC  and  A'  B'  C"  have  the 
side  A  B  and  its  adjacent  angles  equal,  by  construction, 
to  the  side  A^  B'  and  its  adjacent  angles.  Hence,  B  C 
=  B'  a,  and  A  C  =  A^  C  In  like  manner  B  D  = 
B'D',  and  AD  —  A'D'.  Then,  in  the  triangles  A  C  D 
and  A'  C  D^,  we  have  two  sides  in  one,  with  their  included 
angle,  equal  to  two  sides  of  the  other  with  their  included 


POLYEBRONS.  Il5 

angle.  Ilcnco,  the  third  sides  arc  equal,  that  is,  D  C  = 
D'  C.  The  three  sides  of  the  triangle  D  B  C  being  thus 
proved  equal  to  the  three  sides  of  D'  B'  C\  the  angles  are 
equal,  and  the  diedral  angle  on  the  line  A  B  is  equal  to 
that  on  A'  B'.  By  proj^er  changes  in  the  figure  the  same 
may  be  proved  of  the  other  diedral  angles. 

370.  The  two  triedral  angles  mayjn  this  case  be  either 
equal  or  symmetrically  equivalent.  Place  one  plane,  say 
that  of  A  B  C,  and  A'  B'  C  horizontal  with  the  vertices 
from  you  and  th6  lines  A  B,  and  A'  B'  on  your  left.  If, 
now,  the*  lines  A  D  and  A'  D'  are  both  above,  or  both  below, 
the  horizontal  plane,  the  triedral  angles  arc  equal ;  but  if 
one  is  above  and  one  below,  they  are  merely  equivalent, 

371.  A  solid  corner  made  of  several  planes  may  bo 
called  a  polyedral  angle.  If  the  polygon  produced  by  a 
new  ])lane,  cutting  off  a  solid  piece  from  this  corner,  has 
no  reentering  angles,  the  corner  is  called  a  convex  polye- 
dral angl^, 

372.  If  the  sum  of  the  plane  angles  about  a  convex 
polyedral  angle  is  zefo,  the  polyedral  angle  becomes  a 
nee(llo-l>oint,  a  Kne;  and  if  the  sum  of  the  angles  is  2  ti, 
that  is  four  right  angles,  the  polyedral  angle  becomes  a 
plane.    The  sum  is  always,  therefore,  less  than  2  n. 


CHAPTER    III. 

POLYEDRONS. 

•  873.  The  least  number  of  planes  that  can  enclose  a 
space  is  four.  The  solid  thus  enclosed  has  four  triangular 
faces,  and  is  called  a  tetraedron. 

374.   If  two  tetraedrons  have  each  a  solid  angle  enclosed 
in  three  triangles,  equal  and  similarly  arranged  in  one  and 


116  POLYEDRONS. 

in  the  other,  the  tetraedrons  are  equal.  For,  if  one  solid 
angle  be  imagined  laid  in  the  other,  so  as  to  have  one  of 
the  three  triangles  in  one  coincide  with  the  corresponding 
triangle  in  the  other,  the  other  two  will  coincide,  by  Art. 
369  and  by  hypothesis ;  and  the  boundaries  of  the  fourth 
triangle  in  each  thus  coinciding,  the  fourth  triangles  them- 
selves will  coincide.  The  entire  surface  of  one  solid  thus 
coinciding  with  that  of  -the  other,  the  two  solids  are  equal. 

375.  If  two  triangles  in  one  tetraedron  are  equal  to  two 
in  another,  and  similarly  disposed,  and  enclose  the  same 
diedral  angle,  then  the  two  tetraedrons  are  equal.  For  it 
is  manifest  that  the  two  triangles  of  the  one  may  be  ima- 
gined superimposed  upon  the  two  triangles  of  the  other, 
and  will  coincide.  Two  sides  of  each  of  the  unknown 
triangles  in  one  tetraedron  will  then  coincide  with  two 
sides  in  the  unknown  triangles  of  the  other,  and  thus  the 
whole  surfaces  will  coincide. 

376.  Polyedrons,  like  polygons,  are  called  similar  when 
their  homologous  angles  are  equal  and  their  homologous 
sides  are  proportional.  It  follows,  by  induction  from  the 
preceding  sections,  that  polyedrons  are  similar  when  their 
homologous  faces  are  similar  polygons,  similarly  arranged. 

377.  Two  tetraedrons  are  similar  if  a  triedral  angle  in 
one  and  its  homologous  angle  in  the  other  are  composed 
of  similar  triangles,  similarly  arranged.  For,  if  these  two 
angles  are  superimposed,  they  will  coincide,  by  Arts.  369 
and  374,  and  the  fourth  planes  will  be  parallel  to  each 
other.  Hence  follows,  by  Arts.  360  and  374,  the  similarity 
of  the  fourth  triangles,  and  the  equality  of  ratios  in  the 
homologous  sides. 

378.  It  will  also  be  easy  to  show  that,  if  two  triangles 
in  one  tetraedron  are  similar  to  two  in  another,  and  simi- 
larly arranged,  and  enclose  an  equal  diedral  angle,  the  two 
tetraedrons  are  similar. 

379.  If  all  the  planes  of  a  polyedron  except  one  have  a 


POLYEDRONS.  117 

common  point  of  intersection,  tlie  polycdron  is  called  a 
pyramid ;  the  common  point  of  intersection  is  called  the 
vertex  of  the  pyramid ;  the  face,  which  does  not  reach 
the  vertex,  is  called  the  base. 

380.  A  pyramid  is  called  triangular,  quadrangular,  &c., 
from  the  shape  of  its  base.  The  other  faces  are,  of  course, 
always  triangles. 

381.  A  tetraedron  is,  therefore,  a  triangular  pyramid, 
any  face  of  which  may  be  taken  as  its  base. 

382.  If  two  faces  of  a  polyedron  are  equal,  and  their 
homologous  sides  are  parallel,  and  if  each  of  the  other 
faces  is  a  plane  joining  a  pair  of  these  parallel  sides,  the 
polyedron  is  called  a  prism.  The  parallel  faces  are  called 
the  bases  of  the  prism.  The  other  faces  are  evidently 
parallelograms.  A  section  parallel  to  the  base  of  a  prism 
is  readily  shown  to  be  a  polygon  equal  to  the  base. 

383.  When  the  bases  of  a  prism  are  parallelograms,  the 
prism  is  called  a  parallelopipedon. 

384.  A  right  parallelopipedon  is  a  prism  of  which 
every  face  is  a  rectangle.  When  each  face  is  a  square, 
the  prism  is  called  a  cube. 

385.  When  a  pyramid  is  intersected  by  a  plane  parallel 
to  the  base,  the  part  intercepted  between  the  bases  is 
called  the  frustum  of  the  pyramid.  The  part  above  the 
cutting  plane  is  easily  shown  to  be  a  pyramid,  Avith  all  its 
angles  equal  to  those  of  the  given  pyramid,  and  therefore 
similar  to  it. 

386.  Let  a  be  the  length  of  one  side  of  the  base  of  a  pyra- 
mid, a'  that  of  the  homologous  line  on  the  upper  end  of  the 
frustum;  /ithe  height  of  the  pyramid,  h'  that  of  the  similar 
pyramid  cut  off;  b  and  h'  the  slant  heights  of  the  pyramids 
on  the  edge  at  the  left  end  of  a  and  a'. 

By  similar  triangles  we  have  a:  a'  =Lh',h',  Also  h  :  h'  = 
h  :  h'.  Whence  h:  h'  =:a:  a^.  Whence,  by  the  theory 
of  proportions,  a  —  a'  laz^h  —  h!  \h.    Thus  the  total 


118  POLYEDKONS. 

dimensions  of  the  pyramid  are  obtained  from  that  of  the 
frustum,  since  h  —  W  \'$>  shnply  the  vertical  height  of  the 

frustum  ;  and  h  = r 

a  —  a 

Example.  What  is  the  height  of  a  pyramid  "whose  base 

has  sides  of  3,  6,  4^,  and  6  inches,  and  at  the  perpendicular 

heiglit  of  tAvo  inches  the  sides  of  the  frustum  are  4,  8,  6, 

and  8  inches  ?     What  is  the  slant  height  on  the  corner,  on 

which  the  shxnt  height  of  the  frustum  is  3  inches? 

387.  Any  polyedron  can  be  divided  into  pyramids  by 
simply  selecting  a  point  within  the  polyedron  for  a  com- 
mon vertex,  and  taking  the  faces  of  the  polyedron  as  bases 
for  the  pyramids.  By  taking  the  common  vertex  for  the 
pyramids  in  the  surface  of  the  polyedron,  the  number  of  the 
pyramids  may  be  reduced.  Thus  a  right  parallelopipedon 
may  be  divided  into  six  pyramids ;  but  by  bringing  the 
common  vertex  up  to  one  of  the  faces,  the  pyramid  of 
which  that  face  was  base  becomes  zero,  and  the  pyramids 
are  reduced  to  five ;  on  moving  the  vertex  to  one  of  the 
edges,  a  second  pyramid  becomes  zero, reducing  the  number 
to  four ;  and  on  taking  a  vertex  of  the  parallelopipedon  as 
the  common  vertex,  the  pyramids  are  reduced  to  three. 

388.  Any  polyedron  can  be  divided  into  triangular 
pyramids  by  simply  dividing  each  base  in  Art.  387  into 
triangles. 

389.  Two  bodies  which  are  composed  of  equal  and  simi- 
larly arranged  triangular  pyramids  are  evidently  equal. 


ABEAS.  119 


CHAPTER   IV. 

AREAS. 

390.  When  two  polygons  have  all  the  angles  of  one 
equal  to  those  of  the  other,  and  similarly  arranged,  and 
their  homologous  sides  proportional,  —  i.  e.,  each  pair  hav- 
ing the  same  ratio  to  each  other,  the  polygons  are  called 
similar. 

391,  Similar  polygons  may  evidently  be  divided  into 
snnilar  triangles  by  diagonals  from  homologous  vertices. 

392  Lines  drawn  in  a  similar  manner,  in  two  similar 
polygons,  may  evidently  be  made  the  sides  of  similar  trian- 
gles, and  shown  to  have  the  same  ratio  as  homologous  sides 
of  the  polygons. 

393.  Hence  the  altitudes  of  similar  triangles  have  the 
same  ratio  as  their  bases. 

394.  Let  h  be  the  base  and  h  the  altitude  of  a  triangle, 
and  X  be  their  ratio  to  the  base  and  altitude  of  a  similar 
triangle.  The  base  of  the  second  triangle  will  then  be 
h  X  and  its  altitude  h  x.  The  area  of  the  first  will  be  J-  h  h^ 
and  of  the  second  ^  hb  x^.  The  ratio  of  these  areas  will 
therefore  be  x"^. 

Calling  now  the  bases  and  altitudes  h  and  B,  h  and  H, 
and  the  areas  s  and  S,  we  have 

B:h  =  ll:h  =  x\ 

^  :  s  zizx^  =:W-.h''  =  ir  :  h\ 

395.  It  may  easily  be  shown,  by  help  of  equation  (10.) 
(Art.  341),  in  the  theory  of  proportions,  that  the  areas  of 
-similar  polygons,  and  of  any  homologous  areas  in  or  about 
similar  polyedrons,  are  in  the  same  ratio ;  in  other  words, 
that  in  similar  figures  homologous  lines  have  all  the  same 


120  AREAS. 

ratio,  and  that  this  ratio  multiplied  by  itself  will  give  the 
ratio  of  homologous  surfaces  ;  or,  in  other  words,  that  in 
similar  figures  homologous  surfaces  are  in  the  ratio  of 
squares  on  homologous  lines,  or  of  circles  on  homologous 
lines  as  diameters. 

396.  The  similar  polyedrons  spoken  of  in  Arts.  395, 
385,  376,  may  be  defined  as  polyedrons  capable  of  being 
divided  into  similar  tetraedrons  similarly  arranged. 

397.  By  the  reasonmg  alluded  to  in  Art.  395,  it  may  be 
shown  that  the  external  surfaces  of  similar  polyedrons  are 
in  the  ratio  of  squares  built  upon  their  homologous  edges, 
or  upon  any  homologous  lines ;  also  that  any  pair  of  ho- 
mologous faces  are  in  the  same  ratio. 

398.  If  any  two  pyramids  be  cut  by  a  plane  passing  at 
equal  distances  from  their  summits,  the  areas  of  the  sec- 
tions have  a  fixed  ratio,  whatever  be  that  distance. 
Proof,  Let  the  plane  pass  first  at  the  distance  A,  and 
secondly  at  the  distance  h'  from  the  summits,  and  the  areas 
of  the  sections  be  in  the  first  case  A  and  B,  in  the  second 
a  and  h.    We  have,  by  Art.  397,  — 

'    K      h?  B      A2 

A      B  '  . 

Hence?  — =t")  and  by  (3.),  in  the  theory  of  proportions, 

we  have  :^=-i7>  which  is  what  we  wish  to  prove. 
B      h 


VOLUMES.  121 

CHAPTER    V. 

VOLUMES. 

399.  The  volume  of  a  solid  is  the  ratio  which  it  bears 
to  a  solid  uuit.  The  solid  unit  generally  emjDloyed  is  a 
cube,  whose  faces  are  units  of  area,  and  wliose  edges  are 
units  of  length. 

^  400.  It  will  readily  be  perceived  that  the  volume  of  a 
right  parallelopiped  is  the  product  of  the  lengths  of  its  three 
edges.  If  each  edge  is  commensurable  with  the  unit  of 
length,  the  right  parallelopiped  may  be  divided  into  cubes 
by  three  series  of  planes  parallel  to  its  faces,  evidently 
equal  in  number  to  the  product  of  the  numbers  into  which 
each  edge  is  divided.  If  the  edges  are  incommensurable, 
we  can  choose  a  unit  as  small  as  we  please,  and  so  multi- 
ply our  planes  that  the  parallelopipedon  shall  need  but  an 
infinitesimal  change  to  render  it  capable  of  being  divided 
into  cubes.  If  the  right  parallelopided  cannot  be  divided 
into  cubic  inches,  it  may  be  into  cubic  tenths,  or  hun- 
dredths, or  thousandths,  &c.,  of  an  inch. 

401.  Any  parallelopipedon  is  equivalent  in  volume  to 
any  other  parallelopipedon  of  equivalent  base  and  equal  al- 
titude. Proof,  Set  the  two  parallelopipeds  upon  one  plane, 
and  move  a  second  plane,  parallel  to  the  bases,  from  above 
steadily  down  until  the  two  planes  coincide.  This  mov- 
ing plane  moves  with  equal  velocity  through  each  paral- 
lelopiped, and  the  sections  of  the  two  are  constantly  equiv- 
alent from  first  to  last.  The  two  solids  therefore  pass 
equal  volumes  through  the  moving  plane  in  equal  times, 
and  the  volumes  of  the  two  are  equal  when  the  two  planes 
coincide. 

402.  Corollary.  Any  parallelopipedon  is  equivalent  to 
a  right  parallelopiped  on  of  equivalent  base  and  altitude. 

11 


122  VOLUMES. 

403.  Let  the  six  planes  of  a  parallelopiped  be  A  and  A', 
B  and  B\  C  and  C ;  similar  letters  denoting  parallel  faces. 
Cut  the  solid  by  a  plane,  A'^,  perpendicular  to  B  and  to  C. 
Transpose  the  parts,  so  that  A  shall  coincide  with  A',  making 
the  surfaces  on  A''  new  bases  for  the  solid.  Cut  this  new 
solid  by  a  plane,  C,  perpendicular  to  A'^  and  to  B,  and 
transpose  so  that  C  shall  coincide  with  C.  The  solid  is  thus, 
without  loss  or  addition,  converted  into  a  right  parallelo- 
piped ;  two  bases,  B  and  B',  remaining,  in  altered  form, 
of  the  same  size  as  before,  and  their  distance  apart,  as 
measured  on  the  intersection  of  A''  with  C,  being  un- 
changed. From  Art.  402  thus  proved.  Art.  401  may  be 
drawn  as  a  corollary.  But  if  the  plane  A"  cuts  A  or  A', 
this  proof  needs  modification,  by  resections. 

404.  The  diagonal  of  B  being  parallel  to  that  of  B\  a 
plane  may  be  passed  through  these  two  diagonals,  dividing 
every  section  made  by  a  plane  parallel  to  B  into  two  equal 
triangles,  and  the  parallelopiped  into  two  triangular  prisms, 
equivalent  to  each  other.  Conversely  a  triangular  prism 
may  be  considered  as  one  half  a  parallelopiped. 

405.  Two  triangular  prisms  of  equivalent  base  and  equal 
altitude  are,  by  Arts.  401  and  404,  equivalent  in  volume. 

406.  The  volume  of  a  parallelopiped,  or  of  a  prism,  is,  by 
Arts.  400,  402,  404,  405,  found  by  multiplying  the  area  of 
its  base  by  its  altitude. 

407.  Pyi-amids  of  equivalent  base  and  equal  altitude  are 
equal.  Proof,  Let  the  pyramids  be  set  upon  a  plane,  and 
a  second  plane,  parallel  to  the  first,  move  steadily  from  the 
vertex  of  the  pyramids  to  their  base.  As  the  sections  of 
the  pyramids  made  by  the  second  plane  are,  by  Art.  398, 
at  each  instant  equivalent,  the  pyramids  must  be  passing 
with  equal  velocity  through  the  second  plane,  and  the  total 
amounts  passed  through  at  any  instant  are  equivalent,  and 
the  whole  amounts,  when  the  two  planes  coincide,  will  be 
equal. 

408.  Every   triangular    prism   is   divisible    into    tlirce 


VOLUMES.  123 

equivalent  tetraeclrons.  Proof.  Lot  the  ends  of  tlie  prism 
be  the  equal  triangles  ah  c  and  a!  h'  (/,  a  and  a!  being  at 
the  ends  of  tlie  same  edge,  etc.  Pass  two  planes,  aV  d 
and  a  h  d^  through  the  prism,  and  you  have  manifestly  di- 
vided it  into  three  triangular  pyramids.  But  of  these,  two 
have  the  common  vertex  a,  and  the  equal  bases  hU  d  and 
hcd^  and  are  therefore  equal ;  while  the  third  has  a  com- 
mon vertex  </,  with  the  first  of  the  others,  and  a  base  a  a'  h' 
equal  to  its  base  a  h  h'^  and  is  therefore  equal  to  either  of 
the  others. 

409.  Corollary,  The  volume  of  a  triangular  pyramid  is 
found  by  multiplying  the  area  of  its  base  by  one  third  its 
altitude. 

410.  Corollary,  The  volume  of  any  pyramid  is  found 
by  multiplying  its  base  by  one  third  its  altitude. 

411.  When  a  prism  is  divided  by  a  plane  not  parallel  to, 
nor  intersecting,  the  ends,  each  part  of  the  prism  is  called 
a  truncated  prism. 

412.  The  volume  of  a  truncated  triangular  prism  is 
equivalent  to  that  of  three  pyramids  having  each  a  base 
equal  to  that  of  the  prism,  and  altitudes  equal  to  the  alti- 
tudes of  the  three  vertices  of  the  prism.  Proof.  Using 
the  notation  of  Art.  408,  consider  d  h'  d  as  the  base,  and 
ah  c  as  the  unequal  triangle  at  the  truncated  end.  It  is 
manifest  that  of  the  three  tetraedrons  into  which  the  planes 
ah'  d  and  ah  d  divide  the  prism,  one,  viz.,  a  a'  h'  c/,  is  one 
of  the  required  three.  A  second,  viz.,  d  ah  h\  may  be 
proved  equivalent  to  a  second  of  the  required  three,  viz., 
to  h  a'  h'  c'.  For  they  maybe  considered  as  having  a  com- 
mon base,  hh' d^  and  their  vertices  a  and  a'  are  in  a  line 
parallel  to  the  plane  of  that  base.  Finally,  the  third  pyra- 
mid, ahcd^  is  equivalent  to  the  third  required  pyramid, 
c  a'  h'  c^,  because  their  vertices  a  and  a'  are  at  equal  alti- 
tudes from  the  bases  hcd  and  h'  c c',  and  those  bases  are 
equivalent  triangles  having  the  side  c  d  in  common,  and 
the  vertices  h  and  V  in  a  line  parallel  to  it. 


124  THE     CONE. 

413.  The  volumes  of  similar  triangular  pyramids  are 
proportioned  to  the  cubes  built  upon  homologous  lines. 
Proof,  Let  h  be  one  edge  of  the  base,  h  the  altitude  of 
that  base  as  a  triangle,  and  I  the  altitude  of  the  pyramids. 
For  a  similar  pyramid  these  lines  become  h  ir,  li  cc,  and  I  x. 
The  areas  of  the  bases  will  be  J  5  A  and  ^h  h  x^.  The 
volumes  will  he  ^b  hi  and  ^h  hlx^. 

Using  5,  B,  A,  H,  /,  L,  v,  and  V,  as  in  Art.  394,  we  have 

That  is,  the  volumes  are  in  the  ratio  of  the  cubes  of  homol- 
ogous lines;  or  the  ratio  of  the  volumes  is  the  third 
power  of  that  of  the  lines. 

414.  Any  polyedron  being  decomposable  into  triangular 
pyramids,  the  last  theorem  may  be  extended  to  any  similar 
polyedral  figures. 


CHAPTER   VI. 

THE   CONE. 

415.  If  one  point  in  a  straight  line  be  held  fast,  while 
the  line  turning  freely  on  that  point  be  caused  to  glide 
through  a  plane  curve,  the  line  passing  freely  in  space 
marks  out  a  surface  called  the  surface  of  a  cone,  whose 
vertex  is  the  fixed  point. 

416.  When  the  curve  is  a  circle,  and  a  line  through  its 
centre,  perpendicular  to  its  plane,  passes  through  the  vertex 
of  the  cone,  the  cone  is  called  a  circular  cone ;  and  this  is 
the  cone  usually  spoken  of  as  the  cone.  The  right  line 
through  the  vertex  and  the  centre  of  the  circle  is  called 
the  axis  of  the  cone. 


THE    CONE.  125 

417.  The  base  of  a  cone  is  the  plane  figure  intercepted 
by  its  surfiice  on  any  plane  which  cuts  entirely  across  it. 

418.  The  two  parts  of  a  cone  lying  on  opposite  sides  of 
its  vertex  are  called  its  nappes.  In  ordinary  geometry  we 
confine  our  attention  to  one  nappe,  and  to  the  part  inter- 
cepted between  the  vertex  and  the  base. 

419.  As  any  curve  may  be  considered  a  polygon  of  an  in- 
finite number  of  sides,  a  cone  may  be  considered  a  pyra- 
mid, with  a  polygon  for  its  base. 

420.  Hence  the  volume  of  a  cone  is  one  third  the  area 
of  its  base,  multiplied  by  the  altitude  of  its  vertex  above 
the  base. 

421.  Hence,  also,  sections  by  planes  parallel  to  the  base 
will  be  figures  similar  to  the  base.  The  altitude  of  such  a 
section  above  the  base,  and  the  lengths  of  two  homologous 
lines  in  the  section  and  the  base,  give,  by  Art.  386,  data  to 
determine  the  altitude  of  the  cone. 

422.  When  a  circular  cone  has  a  circular  base,  it  is  called 
a  right  cone. 

423.  The  right  cone  may  be  imagined  as  generated  by 
the  revolution  of  a  right  triangle  about  one  leg  as  axis. 

424.  The  convex  surface  of  a  right  cone  may  be  con- 
ceived as  composed  of  infinitesimal  triangles,  all  with  a  com- 
mon vertex  (that  of  the  cone),  and  all  with  equidistant 
bases,  forming  the  circumference  of  the  base  of  the  cone ; 
so  that  this  surface  is  measured  by  the  product  of  half  this 
circumference  into  the  length  of  the  side. 

425.  Let  r  be  the  radius  of  the  base  of  a  right  cone,  and 
then  2  TT  r  will  be  its  circumference.  Let  I  be  its  altitude, 
and  then  V  r  ^  -}~  ^^  yfS}^^  by  the  Pythagorean  proposition,  be 
its  slant  height.  We  have,  therefore,  for  the  area  of  the 
base,  s  zz:  TT  r  2,  and  for  the  area  of  the  convex  surface  S  = 
n  T  ( ^^  +  ^^ )  ^ ;  also,  the  volume,  v  ^=z  ^n  r'^  L 

426.  Pyramids  and  cones  having  equal  altitudes  will 
evidently  be  proportioned  in  their  volumes  to  their  bases. 

11* 


126  OF    THE    SrilERE. 

427.  When  the  vertex  of  a  cone  is  infinitely  distant 
from  the  base,  the  cone  is  called  a  cylinder;  and  if  the  cone 
is  a  right  cone,  the  cylinder  is  a  right  cylinder.  But  in 
the  consideration  of  a  cylinder  we  usually  limit  ourselves 
to  a  part  enclosed  between  parallel  bases.  Thus,  a  cylinder 
is  simply  a  prism,  with  its  parallelogram  faces  infinitely 
numerous  and  infinitely  narrow. 

428.  The  sections  of  a  cylinder  by  planes  parallel  to  its 
base,  are  evidently  figures  equal  to  the  bases. 

429.  If  the  walls  of  the  cylinder  are  perpendicular  to  its 
base,  it  is  evident  that  the  convex  surface  is  measured  by 
the  product  of  the  height  of  the  cylinder  into  the  periph- 
ery of  the  base. 

430.  The  volume  of  the  cylinder  of  Art.  429  is  evidently 
the  product  of  the  area  of  the  base  by  the  altitude  of  the 
cylinder. 

431.  Hence,  the  volume  of  a  cone  is  one  third  that  of  a 
cylinder  of  equivalent  base  and  equal  altitude. 


CHAPTER    VII. 


OF  THE    SPHERE. 


432.  A  SURFACE  which  curves  equally  in  all  directions, 
for  all  distances,  from  any  initial  point,  is  called  the  sur- 
face of  a  sphere. 

433.  There  is  a  point  within  a  sphere  equally  distant 
from  all  parts  of  its  surface,  which  may  be  called  the  centre 
of  the  sphere.  For,  by  definition,  if  we  run  round  the  sphere 
in  any  direction,  keeping  in  the  plane  in  which  we  start, 
perpendicular  to  the  curved  surface,  we  make  a  circle  of  a 
given  size.  Moreover  the  planes  of  these  circles  are  per- 
pendicular to  the  tangent  plane  at  the  initial  point.   Hence, 


OF    THE    SPHERE.  127 

their  diameters  coincide,  and  they  liavc  a  common  centre 
—  the  centre  of  tlic  sphere. 

434.  The  sphere  may  be  conceived  as  generated  by  tlie 
revohition  of  a  semicircle  about  its  diameter  as  axis. 

435.  The  words  diameter  and  radius  may  be  used  of  a 
s])here  in  a  sense  analogous  to  that  in  which  tliey  are  used 
of  a  circle. 

436.  All  radii  of  a  sphere  arc,  then,  equal. 

437.  Every  section  of  a  sphere  made  by  a  plane  is  a 
circle.  Proof.  Let  fall  from  the  centre  a  perpendicular 
upon  the  plane.  Join  the  centre  to  all  parts  of  the  inter- 
section of  the  surface  with  the  plane  by  straight  lines.  As 
these  lines  are  equal,  they  strike  the  plane  at  equal  dis- 
tances from  the  foot  of  the  perpendicular.  That  foot  is 
therefore  the  centre,  and  that  line  of  intersecting  surfaces 
the  circumference  of  a  circle. 

438.  When  the  plane  passes  through  the  centre  of  the 
sphere,  the  circle  is  called  a  great  circle.  All  great  circles, 
in  the  same  sphere,  are  manifestly  equal.  All  other  circles 
on  the  sphere  are  called  small  circles. 

439.  A  spherical  triangle  is  a  spherical  surfjicc  enclosed 
between  three  arcs  of  great  circles. 

440.  The  sides  of  the  spherical  triangle  measure  the 
plane  angles  of  the  triedral  angle  made  by  their  planes  at 
the  centre  of  the  sphere.  They  are  therefore  not  expressed 
in  units  of  length,  like  the  sides  of  plane  triangles,  but  in 
degrees  and  minutes,  which  express  simply  their  ratio  to 
the  circumference  of  the  sphere. 

441.  The  angles  of  a  spherical  triangle  are  the  same  as 
the  diedral  angles  made  by  the  planes  of  their  sides. 

442.  Any  two  great  circles  bisect  each  other;  for  the 
intersection  of  their  planes  must  pass  through  the  centre 
of  the  sphere,  and  be  a  diameter  in  each  circle. 

443.  Any  side  of  a  spherical  triangle  is,  by  Arts.  440 
and  368,  less  than  the  sum  of  the  other  two. 


128  OF   THE    SPHERE. 

444.  The  shortest  path  from  point  to  point  on  the  sur- 
face of  a  sphere  is  the  arc  of  a  great  circle. 

This  is  assumed  as  axiomatic  in  Arts.  432,  433.  But  if 
we  assume  other  definitions  of  a  sphere,  we  may  prove 
this  proposition  thus  :  — 

(1.)  Two  great  circles,  having  two  points  of  their  circum- 
ferences in  common,  coincide ;  for  these  two  points  and 
the  centre  fix  their  planes  as  coincident. 

(2.)  A  great  circle  and  small  circle  cannot  coincide  for  a 
finite  distance ;  for,  if  two  arcs  coincide,  their  planes  must 
coincide,  and  their  radii  be  equal,  and  the  circles  be  one 
and  the  same. 

(3.)  Let  a  and  h  be  points  on  the  surface  of  a  sphere,  con- 
nected by  the  arc  a  5  of  a  great  circle,  and  the  arc  a  rah 
of  a  small  circle.  Draw  the  arcs  of  great  circles  a  m  and 
TYh  h.  Then,  by  Art.  443,  ab<^am^mb.  But  by  draw- 
ing arcs  of  great  circles  from  a,  w,  and  ^,  to  intermediate 
points  on  the  arc  of  the  small  circle,  it  may  be  shown  that 
a  m  '\'  mhi^  less  than  the  sum  of  the  four  arcs ;  and  by 
redivision,  that  these  four  are  less  than  the  sum  of 
eight  arcs  of  great  circles,  placed  as  consecutive  chords  in 
the  small  circle.  This  process  may  be  continued  until  the 
polygon  is  undistinguishable  from  the  arc  of  the  small 
circle.  A  similar  process  may  evidently  be  applied  to  any 
curve,  connecting  a  and  ^,  other  than  the  arc  a  h, 

445.  The  sum  of  the  sides  of  a  convex  spherical  poly- 
gon can  evidently  not  exceed  360°,  i.  e.,  the  circumfer- 
ence of  a  great  circle. 

446.  The  sum  of  the  angles  of  a  spherical  triangle  can-* 
not,  by  Art.  441,  exceed  3  tt,  or  be  less  than  tt. 

447.  When  the  sum  of  the  angles  of  a  spherical  triangle 
is  increased  to  six  right  angles,  the  triangle  becomes  a 
hemisphere. 

448.  When  the  sum  of  the  angles  of  a  spherical  triangle 
is  decreased  to  two  right  angles,  the  triangle  becomes  infin- 


OP   THE    SPHERE.  129 

itesimal  in  comparison  with  the  spliere.  Thus  the  triangles 
of  ordinary  surveying  are  considered  as  plane  triangles, 
infinitesimal  in  comparison  with  the  sui-ftice  of  the  earth ; 
but  the  triangles  in  surveying  for  maps,  with  sides  of  miles 
in  length,  are  spherical  triangles,  the  sum  of  whose  angles 
exceeds  tt, 

449.  If  a  diameter  of  the  sphere  be  drawn  perpendicu- 
lar to  the  plane  of  a  circle,  it  passes  through  the  centre  of 
the  circle,  and  its  extremities  are  called  poles  of  the  circle. 

450.  The  poles  of  a  circle  may  be  called  its  surface 
centres,  and  the  arcs  of  great  circles  from  the  pole  to  the 
circumferences  may  be  called  surface  radii. 

451.  The  angle  of  two  arcs  of  great  circles  (being  in  fact 
a  diedral  angle)  may  either  be  measured  by  the  angle  made  , 
by  the  intersections  of  the  planes  of  the  circles  with  the 
l)lane  tangent  to  the  sphere  at  the  vertex  of  the  two  arcs 
(as  is  done  in  geodetic  surveying),  or  by  an  arc  struck 
with  a  surface  radius  of  90°,  and  the  vertex  as  pole,  and 
intercepted  between  the  two  arcs,  prolonged  if  need  be. 

452.  A  plane  perpendicular  to  the  end  of  a  radius  is 
evidently  a  tangent  plane. 

453.  With  the  vertices  of  a  triangle  as  centres,  and  with 
radii  each  of  90^,  draw  three  arcs  intercepting.  The  trian- 
gle thus  formed  is  called  the  polar  triangle  of  the  first. 

454.  Each  vertex  of  the  polar  triangle  is  by  hypothe- 
sis at  90°  from  each  extremity  of  one  side  of  the  original 
triangle,  and  is,  therefore,  the  pole  of  that  side.  For  the 
great  circle  with  that  pole  would  pass  through  those 
extremities,  and  therefore  coincide  with  that  side.  The 
original  triangle  is,  therefore,  a  polar  triangle  to  its  polar 
triangle  ;  and  the  two  triangles  may  simply  be  called  polar 
triangles. 

455.  If  the  arcs  of  Art.  453  be  limited  so  that  the  pole 
of  each  arc  be  on  the  same  side  of  it,  as  the  triangle  is, 
then  the  sides  of  a  triangle  are  supplements  to  the  angles 


130  •  OF   THE    SPHERE. 

of  its  polar  triangle.  Proof,  Let  a  he  and  A  B  C  be  the 
vertices  of  the  two  triangles.  Let  h'  d  be  the  points  at 
which  a  h  and  a  c,  prolonged  if  necessary,  intercept  B  C. 
Now  we  have,  by  construction,  — 

Be' =  90°,  Q>h'=.^^. 
But  Bc'znB^'  +  ^/c',  and  B  5'  + C  ^>' zz:  B  C. 

Hence,    ^h> -\-h' d -^-Qh' =  180°  =  V>Q-^h'  d. 
But  y  d  is  the  measure  of  the  angle  a.     Hence,  the  angle 
a  and  the  side  B  C  are  supplements  of  each  other ;  their 
sum  is  TT  HZ  180°. 

456.  If  the  three  sides  of  a  triangle  are  respectively 
equal  to  the  three,  sides  of  another,  the  two  triangles  are 
said  to  be  equilateral  with  respect  to  each  other. 

457.  If  two  triangles  on  the  same  sphere,  or  on  equal 
spheres,  are  equilateral  with  respect  to  each  other,  they 
are  also  equiangular  with  each  other ;  which  follows  from 
Arts.  440  and  369. 

458.  Place  the  centres  of  the  equal  spheres  together, 
and  bring  one  side  of  one  triangle,  say  A  B,  into  coinci- 
dence with  the  equal  homologous  side  A'  B'  on  the  other 
Iriangle.  ^  If,  now,  the  equal  angles  A  and  A'  lie  on  the 
same  side  of  AB,  the  triangles  evidently  coincide.  But 
if  the  angles  A  and  A'  lie  on  opposite  sides  of  the  com- 
mon side,  these  triangles,  equilateral  and  equiangular  with 
respect  to  each  other,  are  called  symmetrical. 

459.  If  two  triangles  are  equiangular  with  respect  to 
each  other,  they  are  also  mutually  equilateral.  Proof,  For 
their  polar  triangles  are,  by  Art.  455,  equilateral,  and  there- 
fore, by  Art.  457,  equiangular  with  respect  to  each  other. 
Hence,  by  Art.  455,  the  triangles  themselves  are  equilateral 
with  respect  to  each  other. 

460.  Spherical  triangles,  having  one  side  and  the  adja- 
cent angles,  or  two  sides  and  the  included  angle,  in  one, 
equal  to  the  like  parts  in  the  other,  are  either  equal  or 
symmetrical.     For  it  may  readily  be  shown  that  one  would 


OP    THE    SPHlERE.  131 

coincide  with  the  other,  or  with  a  triangle  symmetrical 
with  the  other. 

461.  In  measuring  spherical  surfaces  a  peculiar  unit  is 
sometimes  used,  namely,  the  surface  of  a  triangle  whose 
tliree  sides  are  90°,  90°,  and  1°.  This  surface  is  called  one 
degree  of  surface,  and  is  jij^th  of  the  surface  of  a  sphere. 

462.  The  surface  included  between  two  great  semicir- 
cles is  called  a  lune. 

463.  The  surface  of  a  lune  is  double  the  angle  of  the  lune. 
lu  other  words,  the  surface  of  the  lune  is  to  that  of  the 
sphere  as  the  angle  of  the  lune  is  to  360°,  or  double  the 
angle  is  to  720°. 

464.  Symmetrical  spherical  tnangles  are  equivalent  in 
area.  Proof,  Place  the  vertex  of  one  angle  upon  the 
vertex  of  the  equal  angle  in  the  other  triangle,  giving  the 
sides  the  same  direction.  Part  of  the  triangles  will 
coincide.  The  non-coincident  parts  will  be  new  triangles, 
mutually  equiangular,  and  therefore  symmetiically  equi- 
lateral. The  same  operation  may  be  repeated  upon  them, 
and  upon  their  *  non-coincident  parts,  until,  finally,  the 
non-coincident  parts  are  infinitesimal  symmetrical  trian- 
gles, or  zeros,  mutually  equiangular  and  equilateral,  whose 
difference  will  differ  nothing  from  zero. 

465.  To  measure  the  surface  of  a  spherical  triangle.  — 
Solution.  Prolong  one  side,  say  A  C,  into  a  great  circle. 
I'rolong  A  B  into  a  semicircle  meeting  A  C  prolonged  in 
A';  and  in  like  manner  prolong  CB  to  C 

Now,  the  triangle  A'  B  C  is  symmetrical  or  equivalent 
to  the  triangle  required  to  complete  the  triangle  ABC 
into  a  lune  with  the  angle  B. 

And  the  sui-face  of  the  hemisphere  (=360°)  is  equal  to 
the  lune  C  A  B  C  plus  the  triangle  A'  B  O,  plus  the  trian- 
gle A'BC.  The  triangle  A' B  C  is,  however,  the  lune 
.\  B  C  A'  minus  the  triangle  ABC.  Substituting  for  each 
lune  the  double  of  its  angle  gives  us, — 


132  OF   THE    SPHERE. 

360°  =  2  C  +  2  B  —  triangle  AB  C  +  2  A  —  triangle  AB  C. 

360°  z=:2(A  +  B  +  C)— 2  triangle  ABC. 

180°  =  A  +  B  +  C  —  triangle  ABC. 

Triangle  ABCznA  +  B  +  C  — 180°. 

That  is,  the  surface  of  a  triangle  is  to  that  of  its  sphere 

as  the  excess  of  the  sum  of  its  angles  over  180°  is  to  720°. 

466.  To  measure  the  surface  of  afrustutn  of  a  right 
coue.  —  Let  S  be  the  area  of  the  curved  surface  of  the 
frustum,  S"  that  of  the  cone  required  to  fill  the  deficiency 
of  the  frustum,  S'  that  of  the  cone  thus  completed ;  and 
let  5,  s"^  and  s'  be  the  slant  heights  of  the  three  bodies. 
Also,  let  r  and  r'  be  the  radii  of  the  two  bases  of  the  frus- 
tum. We  then  have,  by  Art.  424,  and  by  the  evident 
relations  of  the  bodies,  — 

S'  =  ^  r  s',     S"  zzinr'  s", 

^=:^'  — ^"  =  71  {rs'  —  r'  s"). 

But,  on  jDassing  a  plane  through  the  apex  of  the  cone  and 

the  centre  of  the  base,  it  will  be  evident,  from  similarity 

of  triangles,  that 

r\r'=:s'\s"\  whence,  r'  s'  z=r  s^' ; 
and  it  cannot  alter  the  value  of  any  quantity  to  add  to  it 
r'  s'  —  r  s".    We  may,  therefore,  write,  remembering  that 
s=is'  —  s", 

S  =171  (r  s'  —  r'  s"  -\-r'  s'  —  r  s'^ )  r=  tt  (  r  -f-  ^' )  ^' 
The  quantity  7t(r -^r')  is  the  circumference  of  a  circle, 
whose  radius  is  one  half  the  sum  of  the  radii  of  the  bases ; 
and  it  may  readily  be  shown,  by  Art.  421,  and  reasoning 
similar  to  that  of  Art.  364,  that  this  circle  is  a  section  of 
the  frustum,  parallel  to  the  bases,  bisecting  the  distance 
between  them.  Hence,  the  convex  surface  of  the  frustum 
of  a  right  cone  is  measured  by  the  product  of  its  slant 
height  into  the  circumference  of  a  section  midway  between 
the  bases. 

467.  Pass  a  plane  through  the  centres  of  the  bases  of 
the  frustum,  and  calling  the  length  of  half  the  sections  of 


OP   TOE    SPHERE.  133 

the  bases  r  and  r',  as  in  the  preceding  article,  and  the  sec- 
tion of  convex  side  5,  let  fall  a  perpendicular  of  the  length 
h  from  the  extremity  of  r'  upon  r.  From  the  mid-point 
of  s  draw  a  perpendicular  to  the  axis  of  the  cone,  and  its 
length  is  readily  shown  to  be  ^  (  r  -f-  r' ).  From  the  same 
point  draw  a  perpendicular  to  5,  until  it  touches  the  axis 
of  the  cone,  prolonged  if  need  be,  and  call  the  length  of  this 
line  R.  We  have  now  two  right  triangles,  of  which  the 
sides  h  and  s  in  one  are  respectively  perpendicular  to  two 
sides  i  (>'  +  ^')  ^^^  K  i^^  the  other.  They  therefore 
enclose  equal  angles,  and  the  two  right  triangles  are  simi- 
lar, which  gives 

S  =  2  TT  R  A. 

That  is,  the  convex  surface  of  a  frustum  of  a  right  cone  is 
measured  by  the  product  of  its  altitude  into  the  circum- 
ference of  a  circle  whose  radius  is  a  perpendicular  to  the 
slant  surface  reaching  from  its  mid-point  to  the  axis  of  the 
cone. 

468.  A  portion  of  a  sphere  enclosed  between  two  paral- 
lel planes  — that  is,  a  spherical  segment  with  parallel  bases 
—  may,  if  the  bases  are  brought  infinitely  near  together,  be 
considered  as  an  infinitesimal  frustum  of  a  right  cone, 
whose  axis  passes  through  the  centre  of  the  segment  and 
of  the  sphere,  and  whose  slant  surface  is  perpendicular  to 
the  radii  of  the  sphere  touching  the  surface. 

469.  A  hemisphere  may  thus  be  cut,  by  an  infinite  num- 
ber of  planes  parallel  to  its  base,  into  an  infinite  number  of 
frustums  of  different  cones.  But  all  these  frustums  agree  in 
having  their  axis  pass  through  the  centre  of  the  sphere, 
and  in  having  also  perpendiculars  to  their  slant  suiface 
pass  through  the  centre  of  the  sphere.  Moreover,  the  sum 
of  their  altitudes  is  the  radius  of  the  sphere.  Hence,  by 
Art.  467,  the  surface  of  a  hemisphere  with  the  radius  R  is 

S  =  27rR2. 
12 


134  PEOBLEMS    A^B    THEOREMS. 

But  as  the  area  of  the  base  is  2  nR  x  i^  ^=  ^  K>\  it 

follows  that  the  convex  surface  of  a  hemisiDhere  is  double 
the  plane  surface  of  its  base. 

470.  The  surface  of  a  sphere  is  therefore  equivalent  to 
that  of  four  great  circles. 

471.  As  a  sphere  may  evidently  be  divided  into  pyra- 
mids, with  their  common  apex  at  the  centre,  and  the  sum 
of  their  bases  constituting  the  surface  of  the  sphere,  we 
have,  for  itssoHdity,  by  Art.  410, — 

472.  Using  D  =  2  R  as  the  diameter  of  the  circle  and 
sphere,  we  have  for  the  area  of  a  circle,  — 

A  =  7rR2  =  i7rD2=:  .7854  X  D 2. 
V  =  |7rR3=:j7rD3  =  .5236  X  D^ 


CHAPTER    VIII. 

PROBLEMS  AND   THEOREMS. 

473.  Given  the  angle  which  a  straight  line  makes  with 
a  plane.  Find,  by  geometrical  construction,  the  altitude 
from  the  plane  of  a  point  in  the  line  at  a  given  distance 
from  its  intersection  with  the  plane.  For  example,  what 
is  the  height  of  one  end  of  a  yardstick,  the  other  end  rest- 
ing on  the  floor  at  an  angle  of  37°  ? 

474.  Find,  by  geometrical  construction,  the  angle  which 
a  line  makes  with  a  plane,  having  given  its  altitude  above 
two  given  points  in  the  plane,  vertically  under  it ;  and  find 
the  place  of  its  intersection  with  the  plane.  For  instance, 
let  two  posts  on  level  ground  be  seven  feet  apart,  and  be, 
one  four,  the  other  six  feet  high.  What  is  the  inclination 
to  the  horizon  of  a  line  joining  their  summits,  and  where 
would  it  strike  the  ground  ? 


PROBLEMS  AND  THEOREMS.  135 

475.  Given  the  altitude  of  a  second  plane  above  three 
given  points  in  a  first  plane,  to  find  by  geometrical  con- 
struction the  line  of  intersection  of  the  planes,  and  their 
diedral  angle. 

476.  A  right  line  drawn  from  one  vertex  of  a  parallelo- 
pipedon  to  the  vertex,  which  has  no  plane  in  common 
with  the  first,  is  called  a  diagonal  of  the  parallelopip- 
edon.  Prove  that  a  plane  containing  one  diagonal  may- 
be rotated  upon  it  until  it  contains  a  second.  Prove  that 
all  four  diagonals  have  a  common  point  of  intersection. 
Prove  that  each  diagonal  is  bisected  at  this  point. 

477.  The  convex  surface  of  the  frustum  of  a  sphere  is 
called  a  zone.  Prove  that  the  area  of  a  zone  is  the  prod- 
uct of  the  altitude  of  the  frustum  into  the  circumference 
of  a  great  circle. 

478.  Prove  that  the  two  tangents  from  a  given  point  to 
a  circle  are  equal.  Prove  that  a  right  line  from  the  given 
point  to  the  centre  of  the  circle  bisects  the  angle  between 
the  tangents. 

479.  Prove  that  when  a  chord  is  bisected  by  a  diameter, 
the  semicbord  is  a  mean  proportional  between  the  segments 
of  the  diameter.  Find  a  mean  proportional  between  two 
given  lines. 

480.  Prove  —  what  is  assumed  in  Art.  468  —  that  the 
line  of  tangency  of  surfaces,  when  a  sphere  is  enclosed  in  a 
hollow  cone,  is  the  circumference  of  a  circle  whose  plane 
is  at  right  angles  to  the  axis  of  the  cone. 

481.  Prove  that,  if  two  chords  are  prolonged  until  they 
meet,  the  entire  lines  thus  produced  are  inversely  propor- 
tioned to  the  parts  without  the  circle.  Calling  these  entire 
lines  secants,  prove  that  if,  from  a  given  point,  a  tangent 
and  a  secant  be  drawn  to  a  circle,  the  tangent  will  be  a 
mean  proportional  between  the  secant  and  the  part  with- 
out the  circle. 

482.  Find  the  centres,  by  construction,  of  the  two  oir^ 


136  PROBLEMS  AND  THEOREMS. 

cles  whose  circumferences  pass  through  two  given  points 
and  are  tangent  to  a  given  right  line  in  one  plane  with 
them.    When  would  this  problem  become  absurd  ? 

483.  Find  the  centres  of  the  two  circles  whose  circum- 
ferences pass  through  one  given  point  and  are  tangent  to 
two  given  right  lines  all  in  one  plane.  Remember  that  a 
right  line  passing  through  the  centres  is  readily  found,  and 
that  a  perpendicular  upon  it  from  the  given  point  is  a 
semi-chord  in  both  circles. 

484.  Find  the  centres  of  the  four  circles  which  are  tan- 
gent to  three  given  right  lines  in  one  plane.  In  what  case 
would  the  problem  be  impossible  ? 

485.  It  is  evident  that  circles  concentric  with  those  of 
Art.  483  would  pass  equally  near  the  given  points  and  the 
given  right  lines.  Find,  then,  the  centres  of  the  four 
circles  which  are  tangent  to  a  given  circle  and  to  two  given 
right  lines. 

486.  Given  the  radius  of  a  sphere  and  distance  at  which 
a  plane  passes  from  its  centre.  Find  the  radius  of  the  sec- 
tion. When  does  this  radius  equal  that  of  the  sphere  ? 
when  become  zero  ?  and  when  become  impossible  ? 

487.  A  given  sphere  has  its  centre  on  the  axis  of  a  given 
cone,  at  a  given  distance  from  the  vertex.  Find  the.  radii 
of  the  two  circles  made  by  the  intersection  of  the  surfaces. 
In  what  cases  will  they  be  equal  ?  In  what  case  will  the 
two  circles  coincide?  When  will  they  coincide  with  a 
great  circle  ? 

488.  Given  a  great  circle  in  a  sphere,  and  the  radii  of 
two  small  circles  parallel  to  it.  Find  the  hollow  cones 
tangent  to  the  sphere  on  the  small  circles.  Find  also  the 
cones  whose  intersections  with  the  sphere  would  give  the 
circumferences  of  both  circles.  Four  cones  are  required 
in  each  case. 


a 

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J2\J 

§^466 
467 

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AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  50  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.00  ON  THE  SEVENTH  DAY 
OVERDUE. 


•^CT  20  19?' 

-• 

}:| 

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•^^^ 

i 


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THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY^ 


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